Given a Lie group $\mathfrak{G}$ with finite centre and with Lie algebra $\mathfrak{g}$, I am looking at a simple proof that negative definite Killing form implies compactness. This proof is given here. A sketch of the proof is as follows:

1. The author claims that $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is a closed map;
2. The Killing form $K:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R}$ is $\mathrm{Ad}$-invariant, so a negative definite $K$ allows us to define an inner product $-K$such that $\mathrm{Ad}(\gamma)$ is orthogonal wrt $-K$ i.e. $\mathrm{Ad}(\mathfrak{G})\subseteq SO(\dim \mathfrak{g}, -K)$.
3. $SO(\dim \mathfrak{g}, -K)$ being compact, and $\mathrm{Ad}(\mathfrak{G})$ a closed subgroup by 1., we conclude that $\mathrm{Ad}(\mathfrak{G})$ is itself compact;
4. $\mathfrak{G}$, being an $M$-fold cover of $\mathrm{Ad}(\mathfrak{G})$ (where $M$ is finite by dint of the finite centre), is thus also compact.

Crucial to this proof is the assertion that $\mathrm{Ad}(\mathfrak{G})$ is closed in $SO(\dim \mathfrak{g}, -K)$. I can prove this given nondegeneracy of the Killing form, (e.g. with Lemma 1 of G. Hochschild, "Complexification of Real Analytic Groups") but the author of the first document I linked seems to be saying that this is a much more general and well known property of $\mathrm{Ad}$. What am I missing here?: I think I'm making this harder than it should be through overlooking a simple fact. So, as in my title:

When is $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G}))$ a closed map and why?

Edit: It seems that this is not as trivial as I thought. Hence answers less than a full answer are helpful and acceptable to me. For example, interesting counterexamples (showing when $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is not closed) or sufficient conditions for it to be closed (such as semisimplicity of $\mathfrak{G}$).