= Solution
Use the real Lie-algebra convention of the question, with a <gauge covariant derivative> $D_\mu=\partial_\mu+A_\mu$. The <gauge field strength> is its curvature:
$$
\boxed{F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu],\qquad[D_\mu,D_\nu]=F_{\mu\nu}.}
$$
Write $a_\mu=\delta_XA_\mu/\epsilon=-\partial_\mu X+[X,A_\mu]$. To first order in $\epsilon$,
$$
\epsilon^{-1}\delta_XF_{\mu\nu}=\partial_\mu a_\nu-\partial_\nu a_\mu+[a_\mu,A_\nu]+[A_\mu,a_\nu].
$$
Substitute $a_\mu$. The mixed second derivatives of $X$ cancel. The terms containing first derivatives of $X$ cancel in pairs. The remaining terms are
$$
[X,\partial_\mu A_\nu-\partial_\nu A_\mu]+[[X,A_\mu],A_\nu]+[A_\mu,[X,A_\nu]].
$$
The <Jacobi identity> combines the last two into $[X,[A_\mu,A_\nu]]$. Consequently
$$
\boxed{\delta_XF_{\mu\nu}=\epsilon[X,F_{\mu\nu}].}
$$
The transformation is homogeneous even though the connection transformation contains an inhomogeneous derivative term.
For a finite-dimensional <Lie algebra>, define its <Adjoint representation> by $\operatorname{ad}_X(Y)=[X,Y]$ and its <Killing form> by
$$
\kappa(X,Y)=\operatorname{Tr}_{\mathfrak g}(\operatorname{ad}_X\operatorname{ad}_Y).
$$
It is a symmetric <bilinear form> by cyclicity of the <trace>. The <Jacobi identity> gives $\operatorname{ad}_{[Z,X]}=[\operatorname{ad}_Z,\operatorname{ad}_X]$. Set $A=\operatorname{ad}_Z$, $B=\operatorname{ad}_X$, $C=\operatorname{ad}_Y$. Then
$$
\kappa([Z,X],Y)+\kappa(X,[Z,Y])=\operatorname{Tr}([A,B]C+B[A,C])=\operatorname{Tr}(ABC-BCA)=0.
$$
This proves the <invariant bilinear form on a Lie algebra> property, without assuming simplicity or nondegeneracy.
For definiteness use the <Minkowski metric> $\eta=\operatorname{diag}(1,-1,-1,-1)$ and take a real compact <semisimple Lie algebra> as the gauge algebra. Its positive internal metric is $B=-\kappa$. A <Killing-form Yang-Mills Lagrangian> with the coupling absorbed into the connection is
$$
\boxed{\mathcal L=-\frac1{4g_{\rm YM}^2}B(F_{\mu\nu},F^{\mu\nu})=\frac1{4g_{\rm YM}^2}\kappa(F_{\mu\nu},F^{\mu\nu}),\qquad g_{\rm YM}^2>0.}
$$
The spacetime metric is unchanged by internal <gauge transformations>. Hence
$$
\delta_X\mathcal L=\frac{\epsilon}{4g_{\rm YM}^2}\{\kappa([X,F_{\mu\nu}],F^{\mu\nu})+\kappa(F_{\mu\nu},[X,F^{\mu\nu}])\}=0.
$$
Thus <gauge invariance> follows directly from invariance of the <Killing form>. Other overall conventions are possible, but the energy sign must be checked rather than inferred from a prefactor in isolation.
For physical kinetic terms the internal form must be real, nondegenerate and positive definite after choosing the overall sign. If $\mathcal E_i=F_{0i}$ and $\mathcal B_i=\tfrac12\epsilon_{ijk}F_{jk}$, the above convention has Lagrangian $[B(\mathcal E_i,\mathcal E_i)-B(\mathcal B_i,\mathcal B_i)]/(2g_{\rm YM}^2)$ and physical <energy> density
$$
\mathcal H=\frac1{2g_{\rm YM}^2}\sum_i[B(\mathcal E_i,\mathcal E_i)+B(\mathcal B_i,\mathcal B_i)]\ge0.
$$
The canonical <Hamiltonian> density also contains the nondynamical multiplier and a spatial divergence. With $\Pi_i=\mathcal E_i/g_{\rm YM}^2$ and $D_i\Pi_i=\partial_i\Pi_i+[A_i,\Pi_i]$, <integration by parts> using the <invariant bilinear form on a Lie algebra> gives
$$
\mathcal H_{\rm can}=\mathcal H-B(A_0,D_i\Pi_i)+\partial_i B(A_0,\Pi_i).
$$
The <Gauss law constraint in gauge theory> sets $D_i\Pi_i=0$. If the boundary flux vanishes, or the appropriate boundary contribution is included, the integrated physical <Hamiltonian> is the positive <energy> displayed above.
An indefinite internal form would give gauge-field polarizations with opposite kinetic signs. A degenerate form would fail to supply a kinetic term for some directions. The <compactness criterion from the Killing form> says that negative-definiteness of the <Killing form> of a real finite-dimensional algebra is equivalent to compact semisimplicity; thus the pure Killing-form construction selects compact semisimple real forms. One cannot use a complex-bilinear <Killing form> on arbitrary complex field components as if it were a positive Hermitian metric.
This does not prohibit Abelian gauge theories. A compact Abelian factor has zero <Killing form>, so it needs a separately chosen positive <invariant bilinear form on a Lie algebra>, rather than the <Killing form>. More generally an algebra with a <positive invariant metric on a Lie algebra> is compact reductive, namely a direct sum of a compact semisimple algebra and an Abelian center. A noncompact group can also share the same compact <Lie algebra> through global covering choices in an Abelian factor; positivity is a statement about the algebra and internal metric, not by itself a classification of global gauge-group topology. Quantum matter anomalies and global restrictions would require additional input; no matter content is specified here.
For the finite matrix transformation, let $g(x)\in SU(N)$ and regard $g$ as a multiplication operator. The identity $\partial_\mu g^{-1}=-g^{-1}(\partial_\mu g)g^{-1}$ gives
$$
D'_\mu=gD_\mu g^{-1}=\partial_\mu+gA_\mu g^{-1}-(\partial_\mu g)g^{-1}.
$$
Taking commutators of these differential operators cancels the adjacent multiplication operators $g^{-1}g$, yielding
$$
\boxed{F'_{\mu\nu}=gF_{\mu\nu}g^{-1}.}
$$
This argument keeps the derivatives acting on test fields and avoids treating $D_\mu$ as just a matrix.
There is a normalization issue in the printed last paragraph. The standard <Killing form> defined through the adjoint <trace> on $\mathfrak{su}(N)$ is
$$
\kappa_{\rm Kill}(X,Y)=2N\operatorname{tr}_{\mathbb C^N}(XY),
$$
not simply $\operatorname{tr}(XY)$. For example, with $N=2$ and $X=Y=-i\sigma_3/2$, the defining trace is $-1/2$, whereas the adjoint trace is $-2$. The printed <trace> formula can be used as a rescaled invariant form, with the constant absorbed into the gauge coupling; it has exactly the invariance needed here. For either normalization,
$$
\operatorname{tr}(F'_{\mu\nu}F'^{\mu\nu})=\operatorname{tr}(gF_{\mu\nu}F^{\mu\nu}g^{-1})=\operatorname{tr}(F_{\mu\nu}F^{\mu\nu}),
$$
by cyclicity. This proves finite <Yang-Mills gauge transformation> invariance, with the healthy sign chosen for anti-Hermitian gauge fields. The normalization discrepancy is not a failure of <gauge invariance>.
Finally let $g(x)=e^{\epsilon X(x)}=I+\epsilon X(x)+O(\epsilon^2)$. Since $X\in\mathfrak{su}(N)$ is traceless and skew-Hermitian, this exponential lies in $SU(N)$. Expanding the finite formula gives
$$
A'_\mu=A_\mu+\epsilon[X,A_\mu]-\epsilon\partial_\mu X+O(\epsilon^2),\qquad F'_{\mu\nu}=F_{\mu\nu}+\epsilon[X,F_{\mu\nu}]+O(\epsilon^2).
$$
Thus \b[the stated infinitesimal transformation is the derivative of the finite transformation at the identity]. It describes transformations in the identity component; arbitrary global or large <gauge transformations> need not be generated by one globally defined infinitesimal parameter.
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