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I'm confused as to why (failure of) local compactness is an issue. I think one can just prove this directly as follows. It's rather trivial, though, so maybe I'm missing something.

Let $x\in X$ be a point and $K\subseteq X$ a closed subset with $x\notin K$. Find some coordinate chart near $x$, that is, a map $f:E\to X$ which is a homeomorphism onto some open set containing $x$, where $E$ is a Banach space and $f(0)=x$. Now $f^{-1}(K)$ is a closed subset of $E$ which does not contain $0$. Thus there exists $\epsilon>0$ such that $\|k\|\geq\epsilon$ for all $k\in f^{-1}(K)$. Let $B(0,a)$ denote the open ball in $E$ centered at $0$ with radius $a$. Then we simply observe that the following two open sets are disjoint and "separate" $x$ and $K$: $$x\in f(B(0,\epsilon/3))$$ $$K\subseteq X\setminus f(\overline{B(0,2\epsilon/3)})$$ I'm using Hausdorffness to conclude that $f(\overline{B(0,2\epsilon/3)})$ is closed in $X$.

EDIT: As the comments point out, the point here is exactly to show that $f(\overline{B(0,2\epsilon/3)})$ is closed, which in finite dimensions (i.e. locally compact) follows easily from Hausdorfness. So, as of now this argument is incomplete.

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I'm confused as to why (failure of) local compactness is an issue. I think one can just prove this directly as follows. It's rather trivial, though, so maybe I'm missing something.

Let $x\in X$ be a point and $K\subseteq X$ a closed subset with $x\notin K$. Find some coordinate chart near $x$, that is, a map $f:E\to X$ which is a homeomorphism onto some open set containing $x$, where $E$ is a Banach space and $f(0)=x$. Now $f^{-1}(K)$ is a closed subset of $E$ which does not contain $0$. Thus there exists $\epsilon>0$ such that $\|k\|\geq\epsilon$ for all $k\in f^{-1}(K)$. Let $B(0,a)$ denote the open ball in $E$ centered at $0$ with radius $a$. Then we simply observe that the following two open sets are disjoint and "separate" $x$ and $K$: $$x\in f(B(0,\epsilon/3))$$ $$K\subseteq X\setminus f(\overline{B(0,2\epsilon/3)})$$ I'm using Hausdorffness to conclude that $f(\overline{B(0,2\epsilon/3)})$ is closed in $X$.