Actually, my question is a bit more specific: Does every complex semisimple Lie group $G$ admit a faithful finite-dimensional holomorphic representation?

Of course the analogous question for real semisimple Lie groups has a negative answer -- "holomorphic" having been replaced by "continuous", "smooth" or "real analytic" -- with the canonical counterexample being a nontrivial cover of $\mathrm{SL}(2,\mathbb{R})$.

For a connected complex semisimple Lie group $G$ I believe the answer is "YES." The idea is to piggy back off a 'sufficiently large' representation of a compact real form $G_\mathbb{R}$; here by "compact real form" I'm referring specifically to the definition which allows us to uniquely extend continuous finite-dimensional representations of $G_\mathbb{R}$ to holomorphic representations of $G$. I know (e.g. from the proof of Theorem 27.1 in D. Bump's Lie Groups) that such a definition is possible if we require $G$ to be connected (and I'd like to know if it's possible in general).

The details of the argument for connected $G$ are as follows. Consider the adjoint representation $\mathrm{Ad} \colon G \to \mathrm{GL}(\mathfrak{g})$. Since $G$ is semisimple, $\mathrm{Ad}$ has discrete kernel $K$. Consider next the restriction of $\mathrm{Ad}$ to $G_\mathbb{R}$. Observe that the kernel of this map is also $K$, for otherwise its holomorphic extension is different from the adjoint representation of $G$. Thus $K$ is finite, being a discrete, closed subset of a compact space. So by the Peter-Weyl theorem, we can find a representation $\pi_0$ of $G_\mathbb{R}$ that is nonzero on $K$. Extend $\pi_0$ to a holomorphic representation $\pi$ of $G$ and put $\rho = \pi \oplus \mathrm{Ad}$. Notice that $\rho$ is a holomorphic representation of $G$ with kernel $\ker\pi \cap K = 0$, which is what we were after.

Is the above proof correct?