An old result of Iwasawa is that in any connected Lie group $G$, every compact subgroup is contained in a maximal compact subgroup, and all maximal compact subgroups are conjugate.
Let $G$ be a connected Lie group and let $Z$ be a discrete central subgroup. Then $Z$ has an infinite torsion quotient $Z'=Z/B$ (lemma below). Write $H=G/B$. Then $Z'$ is a discrete central subgroup of $G$ and is infinite torsion. Let $K$ be a maximal compact subgroup of $H$. Every finite $F$ subgroup of $Z'$ is contained in a maximal compact subgroup $K'_F$ of $H$; since $F$ is central and $K$ is conjugate to $K'_F$, we deduce that $F\subset K$. Hence $Z'\subset K$, a contradiction.
This would be fine with a proof of Iwasawa's result; the only problem is that I don't know if it relies on the result of finite generation of discrete central subgroups!
Lemma: if $Z$ is an infinitely generated abelian group, then $Z$ has an infinite torsion quotient.
Proof: let $B$ be a maximal free subgroup. Then $Z/B$ is torsion. If $B$ is finitely generated then $Z/B$ is infinite. Otherwise, choose an infinite subset $(e_n)_{n\ge 1}$ in $B$ and replace $e_n$ by $ne_n$ for all $n$, to obtain a smaller free subgroup $B'$, with $B/B'$ infinite torsion; then $Z/B'$ is infinite torsion.
Actually, instead of all the power of Iwasawa's result, it's enough to prove the result using the weaker result: ($*$) for any connected Lie group $G$ and any increasing sequence $(K_n)$ of compact subgroups of $G$, we have $\overline{\bigcup K_n}$ compact.
Possibly this can done by hand (say, without using such things as Levi factors, just more basic Lie theory), I'll think twice.
Edit: here's a proof of the OP's question relying on little (only on the semisimple case, where the center is discrete and finitely generated):
Let $G$ be a counterexample of minimal dimension. By the previous lemma, we can suppose (up to mod out by a discrete normal subgroup) that $G$ has an infinite, discrete torsion central subgroup $Z$. Taking the semisimple case for granted, $G$ is not semisimple. Let $V$ be the closure of the last nontrivial term of the derived series of its radical. Let $p$ be the projection $G\to G/V$. Since $\dim(G/V)<\dim(G)$, we have $\overline{p(Z)}$ compact. Hence its unit component has finite index, and hence some finite index subgroup of $Z$ is contained in the inverse image $H$ in $G$ of $\overline{p(Z)}^\circ$. Hence $H$ is a counterexample; by minimality, we deduce $\dim(H)=\dim(G)$ and hence $H=G$. That is, $\overline{VZ}$ is dense. We have $[V,Z]\subset V\cap Z$, which is a torsion discrete subgroup of the connected abelian Lie group $V$, and hence is finite. Hence $[G,G]$ is contained in $[V,Z]$, which is finite; by connectedness of $G$, we deduce that $G$ is abelian, and in turn this implies $Z$ finitely generated, and a contradiction.
As regards the semisimple case, if by contradiction $G$ is semisimple and $Z$ is an infinitely generated central subgroup, then the quotient $H$ of $G$ by its center is semisimple and not compactly presented. Then one way to get a contradiction is to use that $G$ is quasi-isometric to its symmetric space, which is non-positively curved and hence large-scale simply connected, and for $G$ this means compactly presented, a contradiction. Of course this latter proof relies on some Riemannian and metric material.