= Solution
Use the <derived series of a Lie algebra>
$$
L^{(0)}=L,\qquad L^{(r+1)}=[L^{(r)},L^{(r)}].
$$
A <solvable Lie algebra> has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional <Lie algebra> is abelian, so $[[L,L],[L,L]]=0$. \b[Every two-dimensional complex <Lie algebra> is solvable, with derived length at most two.] This dimension argument actually works over any field with the alternating definition of a <Lie bracket>.
There is a genuine qualification in the last request: the assertion about all <Irreducible Lie algebra representations> is true for finite-dimensional representations. Without that restriction, the <infinite-dimensional simple module for the two-dimensional affine Lie algebra> is a counterexample. Take $L=\mathbb Cx\oplus\mathbb Cy$, with $[x,y]=y$, and let it act on $V=\mathbb C[t]$ by
$$
(xf)(t)=tf(t),\qquad(yf)(t)=f(t-1).
$$
Then $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a <Lie algebra representation>. Any nonzero <submodule> is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb C[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two <polynomials> have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. \b[The literal unrestricted assertion is false.]
Here is a proof of the intended finite-dimensional assertion, including the essential <Lie theorem> argument. We prove that a nonzero finite-dimensional representation $\rho:L\to\operatorname{End}(V)$ of a complex solvable algebra has a common <eigenvector>, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus\mathbb Cx$. Induction gives $v\ne0$ and $\lambda\in K^*$ with $\rho(y)v=\lambda(y)v$ for all $y\in K$. Write $X=\rho(x)$ and suppress $\rho$ on elements of $K$.
Let $v_j=X^jv$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\ldots,v_{m-1}$. The space $W=\operatorname{span}(v_0,\ldots,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively that
$$
yv_j-\lambda(y)v_j\in\operatorname{span}(v_0,\ldots,v_{j-1}).
$$
Thus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda(y)$ in this basis. The <trace> of a <commutator> is zero, so
$$
0=\operatorname{tr}_W[X,y]=m\lambda([x,y]).
$$
Since we work over $\mathbb C$, $\lambda([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous <eigenspace>
$$
V_\lambda=\{w\in V:yw=\lambda(y)w\text{ for every }y\in K\}
$$
is $X$-invariant: $yXw=\lambda(y)Xw+\lambda([y,x])w=\lambda(y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an <eigenvector>. An <eigenvector> of $X|_{V_\lambda}$ is consequently a common <eigenvector> for $L$.
Its span is a one-dimensional <submodule>. If $V$ is irreducible, that span must be all of $V$. Hence \b[every finite-dimensional irreducible representation of a finite-dimensional complex solvable <Lie algebra> has dimension one]. Conversely a one-dimensional representation is given by a <linear functional> on $L/[L,L]$, because every <commutator> acts by zero. Passing repeatedly to quotients also gives the <simultaneous triangularization of a Lie algebra representation>. Neither the <eigenvector> step nor the <trace> argument extends to the infinite-dimensional counterexample above.
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