Is an open subset of a compact subset of a Hausdorff locally convex TVS paracompact? This repeats the title in a more readable way. Take a compact subset $X$ of a Hausdorff locally convex topological vector space and $U$ be an open subset of $X$. Is $U$ paracompact?
 A: No.  In fact, every compact Hausdorff space $K$ embeds in a locally convex topological vector space.  Namely, take the dual $C(K)'$ of the space of continuous functions on $K$ with the weak* topology, and embed $K$ in $C(K)'$ as the functionals given by evaluations at points.  So any non-paracompact locally compact Hausdorff space can be realized as a counterexample to your question.
A: I claim that if $H$ is complete locally convex topological vector space, then every open subset of a compact subset of $H$ is paracompact if and only if every compact subset of $H$ is perfectly normal.
It is well known that every subspace of a space $X$ is paracompact if and only if every open subspace of $X$ is paracompact (see for instance the general topology text by Dugundji for a proof. A proof is not that difficult though.). The hereditarily paracompact locally connected compact spaces however have a simple characterization.
$\mathbf{Theorem}$ A compact locally connected space $X$ is hereditarily paracompact if and only if every connected open subspace of $X$ is a $F_{\sigma}$-set.
$\mathbf{Proof}$ The proof of this fact is based on the fact that a locally compact space is paracompact if and only if it is the union of a disjoint collection of $\sigma$-compact open sets. Since every compact locally connected space is a finite union of disjoint open sets, we only need to prove this fact for connected spaces.
$\leftarrow$ Suppose that $U$ is an open subset of $X$. Then since $X$ is locally connected, the set $U$ is a disjoint union of connected open sets $(U_{i})_{i\in I}$. However, since each $U_{i}$ is a $F_{\sigma}$-set, each set $U_{i}$ is paracompact. Therefore, the set $U$ is paracompact as well.
$\rightarrow$ Suppose now that $U$ is a connected open subset of $X$. Then $U$ is paracompact, so $U$ is a disjoint union of $\sigma$-compact open sets. However, since $U$ is connected, the space $U$ is a $\sigma$-compact, so $U$ is a $F_{\sigma}$-set. $\mathbf{QED}$.
Let $H$ be a locally convex topological vector space. Then $H$ can be given a canonical uniformity since $H$ is an abelian topological group and since $H$ is determined by a collection of seminorms. Therefore, since $H$ has a canonical uniform space structure, we have a notion of a complete locally convex topological vector space.
If $X$ is a complete locally convex topological vector space and $C\subseteq X$ is compact and $D$ is the convex hull of $C$, then $D$ is totally bounded, so $\overline{D}$ is compact and convex. Take note that every convex subset of $X$ is connected and locally connected.
Recall that a topological space $X$ is perfectly normal if and only if it is normal and every open subset is a $F_{\sigma}$-set. 
$\mathbf{Theorem}$ Let $H$ be a complete locally convex topological vector space. Then the following are equivalent.


*

*Every compact subset of $H$ is perfectly normal.

*Every open subset of a compact subset of $H$ is paracompact.
$\mathbf{Proof}$
$1\rightarrow 2$ If $C$ is a compact subset of $H$ and $U\subseteq C$ is an open subset, then since $C$ is perfectly normal, the set $U$ is a $F_{\sigma}$-set, so $U$ is paracompact.
$2\rightarrow 1$. Since a subspace of a perfectly normal space is perfectly normal, 
it suffices to show that for each compact set $C\subseteq H$ there is a compact perfectly normal set $D$ with $C\subseteq D\subseteq H$. Suppose that $C$ is a compact subset of $H$. Then since $H$ is a complete locally convex topological vector space, the closure of the convex hull of $C$ is also compact. Therefore, there is a compact convex space $D$ with $C\subseteq D$.
Without loss of generality, assume that $0\not\in D$. Then let $f:[0,1]\times D\rightarrow H$ be the mapping where $f(r,x)=rx$. Then let $E=f[[0,1]\times D]$. Then $E$ is a compact space.
Let $U\subseteq D$ be an open set. Let $V=E\setminus D$. Then $U\cup V$ is an open subset of $E$. Furthermore, $U\cup V$ is path connected, so $U\cup V$ is connected. Therefore, since $U\cup V$ is paracompact, the set $U\cup V$ is a $F_{\sigma}$-set in $E$. Therefore, $U$ is also a $F_{\sigma}$-set in $D$. We conclude that $D$ is perfectly normal.
$\mathbf{QED}$
