= Solution
\b[The form attached to $V$ is $B_V(x,y)=\operatorname{tr}_V(xy)$.] More generally, for a <Lie algebra representation> $\rho$, the <Trace form of a Lie algebra representation> is
$$
B_V(x,y)=\operatorname{tr}_V(\rho(x)\rho(y)).
$$
The unqualified <Killing form> is the special case of the <Adjoint representation>,
$$
B(x,y)=\operatorname{tr}_{\mathfrak g}(\operatorname{ad}x\operatorname{ad}y).
$$
The distinction matters: a <Trace form of a Lie algebra representation> can be degenerate even when $\mathfrak g$ is <semisimple Lie algebra>[semisimple], for example on the <trivial Lie algebra representation>.
The <Trace form of a Lie algebra representation> is <bilinear map>[bilinear] and symmetric, because $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. It is an <invariant bilinear form on a Lie algebra>:
$$
B_V([x,y],z)=B_V(x,[y,z]).
$$
This follows by expanding both <commutators> and cyclically permuting factors under the <matrix trace>. Equivalently,
$$
B_V([x,y],z)+B_V(y,[x,z])=0.
$$
Its <radical of a bilinear form> is an <ideal of a Lie algebra>, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The <Killing form> is also preserved by every <automorphism of a Lie algebra>, because the corresponding <adjoint operators> are conjugate. On a complex finite-dimensional <Lie algebra>, the <Cartan criterion for semisimplicity> says that the <Killing form> is nondegenerate exactly when the <Lie algebra> is <semisimple Lie algebra>[semisimple]. The <Cartan solvability criterion> says that $\mathfrak g$ is <solvable Lie algebra>[solvable] exactly when $B(\mathfrak g,[\mathfrak g,\mathfrak g])=0$.
We next construct the <sl2 subalgebra associated with a root>. Use the <root-space decomposition>
$$
\mathfrak g=\mathfrak h\oplus\bigoplus_{\beta\in\Phi}\mathfrak g_\beta.
$$
For $u\in\mathfrak g_\beta$, $v\in\mathfrak g_\gamma$, invariance of the <Killing form> gives
$$
(\beta(h)+\gamma(h))B(u,v)=0\quad(h\in\mathfrak h).
$$
Thus $B(\mathfrak g_\beta,\mathfrak g_\gamma)=0$ unless $\beta+\gamma=0$, and $B(\mathfrak g_\beta,\mathfrak h)=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak g$ now implies that $-\alpha$ is a <root> and that $B$ pairs $\mathfrak g_\alpha$ and $\mathfrak g_{-\alpha}$ nondegenerately.
Nondegeneracy of $B|_{\mathfrak h}$ defines a unique $t_\alpha\in\mathfrak h$ by
$$
B(t_\alpha,h)=\alpha(h)\quad(h\in\mathfrak h).
$$
Choose $x\in\mathfrak g_\alpha$ and $y\in\mathfrak g_{-\alpha}$ with $B(x,y)=1$. Their <Lie bracket> lies in the zero <root space>, namely $\mathfrak h$, and
$$
B([x,y],h)=B(x,[y,h])=\alpha(h)B(x,y)=\alpha(h).
$$
Therefore $[x,y]=t_\alpha$.
The essential <nonisotropic root lemma> is that $\alpha(t_\alpha)\ne0$. Suppose instead that it vanished. Then $[t_\alpha,x]=[t_\alpha,y]=0$, so $\langle x,y,t_\alpha\rangle$ would be a <solvable Lie algebra> with <derived algebra> $\mathbb Ct_\alpha$. Apply the <Lie theorem> to its action on $\mathfrak g$ by the <Adjoint representation>. The <commutator> $\operatorname{ad}t_\alpha=[\operatorname{ad}x,\operatorname{ad}y]$ is strictly upper triangular in a suitable <basis>, hence <nilpotent linear map>[nilpotent]. But $t_\alpha\in\mathfrak h$, so the <root-space decomposition> makes $\operatorname{ad}t_\alpha$ diagonalizable. A diagonalizable <nilpotent linear map> is zero. Thus $t_\alpha$ is central in $\mathfrak g$. The <center of a Lie algebra> of a <semisimple Lie algebra> is zero; equivalently a central element lies in the <radical of the Killing form>. This forces $t_\alpha=0$, contradicting $\alpha\ne0$.
Writing $c=\alpha(t_\alpha)$, define
$$
\boxed{X_\alpha=x,\qquad H_\alpha=\frac{2t_\alpha}{c},\qquad Y_\alpha=\frac{2y}{c}.}
$$
The <root-space decomposition> and $[x,y]=t_\alpha$ give
$$
[H_\alpha,X_\alpha]=2X_\alpha,\qquad [H_\alpha,Y_\alpha]=-2Y_\alpha,\qquad [X_\alpha,Y_\alpha]=H_\alpha.
$$
The three vectors are linearly independent because they lie in the distinct summands $\mathfrak g_\alpha$, $\mathfrak h$, and $\mathfrak g_{-\alpha}$. Their span is therefore a copy of the <sl2 Lie algebra>.
\b[The weight lattice consists of the functionals integral on all coroots.] With $H_\alpha$ the <coroot> above, the <weight lattice> is
$$
P=\{\lambda\in\mathfrak h^*: \lambda(H_\alpha)\in\mathbb Z\text{ for all }\alpha\in\Phi\}
=\bigoplus_{i=1}^{\ell}\mathbb Z\omega_i,
$$
where the <fundamental weights> satisfy $\omega_i(H_{\alpha_j})=\delta_{ij}$ for the <simple roots> $\alpha_j$. Here $P$ lies in the real span of the <roots>, viewed inside $\mathfrak h^*$.
The <classification of finite-dimensional sl2 representations> says that every finite-dimensional complex <sl2 Lie algebra> representation is a <direct sum> of irreducibles $L_m$, $m\in\mathbb Z_{\ge0}$, on which the standard $H$ has <eigenvalues> $m,m-2,\ldots,-m$. Restrict any finite-dimensional <Lie algebra representation> of $\mathfrak g$ to each <sl2 subalgebra associated with a root>. If $v$ has <weight> $\lambda$, then $H_\alpha v=\lambda(H_\alpha)v$, so $\lambda(H_\alpha)$ is an integer. Thus every <weight> lies in $P$. The same restrictions show that the commuting simple <coroots> act diagonalizably, justifying the simultaneous <weight-space decomposition>.
\b[For $\mathfrak{sp}(4)$, the roots are $\pm2\varepsilon_1$, $\pm2\varepsilon_2$, and $\pm\varepsilon_1\pm\varepsilon_2$.] Work on $\mathbb C^4$ with the alternating <bilinear form> having matrix
$$
J=\begin{pmatrix}0&I_2\\-I_2&0\end{pmatrix}.
$$
The <symplectic Lie algebra> is
$$
\mathfrak{sp}(4)=\{M:M^TJ+JM=0\}
=\left\{\begin{pmatrix}A&B\\C&-A^T\end{pmatrix}:B=B^T,\ C=C^T\right\}.
$$
Using the <matrix units> $E_{ij}$, take the <Cartan subalgebra>
$$
\mathfrak h=\langle h_1,h_2\rangle,\qquad h_1=E_{11}-E_{33},\qquad h_2=E_{22}-E_{44}.
$$
Define $\varepsilon_i(a_1h_1+a_2h_2)=a_i$. A regular diagonal element of $\mathfrak h$ has centralizer precisely $\mathfrak h$, and every element of $\mathfrak h$ acts diagonalizably. Thus it is a <Cartan subalgebra>. The requested Cartan decomposition is the <root-space decomposition>
$$
\mathfrak{sp}(4)=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathbb CX_\alpha.
$$
Choose <positive roots> $\varepsilon_1-\varepsilon_2$, $2\varepsilon_2$, $\varepsilon_1+\varepsilon_2$, $2\varepsilon_1$. The <symplectic root sl2 triple> are given explicitly by
$$
\begin{array}{c|c|c|c}
\alpha&X_\alpha&H_\alpha&Y_\alpha\\\hline
\varepsilon_1-\varepsilon_2&E_{12}-E_{43}&h_1-h_2&E_{21}-E_{34}\\
\varepsilon_1+\varepsilon_2&E_{14}+E_{23}&h_1+h_2&E_{41}+E_{32}\\
2\varepsilon_1&E_{13}&h_1&E_{31}\\
2\varepsilon_2&E_{24}&h_2&E_{42}
\end{array}
$$
For the negative <root spaces>, use the corresponding $Y_\alpha$. These eight <root vectors>, together with $h_1,h_2$, form a <basis>: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the <matrix unit> identity
$$
[E_{ij},E_{kl}]=\delta_{jk}E_{il}-\delta_{li}E_{kj}
$$
verifies $[X_\alpha,Y_\alpha]=H_\alpha$ for every row. The diagonal differences verify $[H_\alpha,X_\alpha]=2X_\alpha$ and $[H_\alpha,Y_\alpha]=-2Y_\alpha$. Thus each row supplies a <basis> of the required <sl2 subalgebra associated with a root>.
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