In Morris' "Introduction to Arithmetic Groups", exercise $A4\#9$ he asks us to prove the title. Here all Lie groups are closed subgroups of $SL(\ell,\mathbb{R})$ with finitely many connected components, and a simple Lie group is one without connected, closed, normal, nonabelian subgroups. A semisimple Lie group is a product of simple ones.
He gives a hint: "The identity component of the Zariski closure of $Z(G)$ is a connected normal subgroup of $G$", where $\ell \times \ell$ matrices are identified with $\mathbb{R}^n$ and then we can consider the Zariski closure $\overline{Z(G)}^Z \subset SL(\ell,\mathbb{R})$ and take the identity component, call it $H$.
Suppose $G$ is simple and we take the hint for granted. If $H$ is nonabelian, then $H$ is trivial, and $\overline{Z(G)}^Z$ is discrete in $G$. From the chapter, we know Zariski closed subsets have finitely many components, so $H$ is finite index and we're done. This generalizes immediately to products of simple groups. If $H$ is abelian, our assumptions don't seem to buy us anything.
Also, how do we prove what the hint says? I don't see why $\overline{Z(G)}^Z$ is a group, why it should be contained in $G$ as a subset of $SL(\ell,\mathbb{R})$, or why it should be normal. The section where the exercise appears proves that "all subgroups of $SL(\ell,\mathbb{R})$ are almost Zariski closed", which may be relevant. I feel like there's a lot of interesting intuition for how group theory affects topology here but I'm having trouble sorting out the details. Can you fill them in?
$\textbf{Edit:}$ Page $11$ of these notes shows the Zariski closure of a group is again a group, and using a similar strategy I can show the Zariski closure of a normal subgroup is again normal. Morris' definition of 'almost Zariski closed' together implies that if $G$ is connected, $G$ is the identity component of a Zariski closed group, say $C$ and in that case $$(\overline{Z(G)}^Z)^\circ \subset C^\circ = G$$ Where $^\circ$ means identity component. If Zariski closure and identity component commute, then this would show $\overline{Z(G)}^Z \subset G$ in general, and we'd be done.