How do I show that if $A$ is compact and $U \supseteq A$ is open, then there is an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq U$?

Question

This question is from Wayne Patty's Topology Section 5.2.

Consider $A$ be a compact subset of a regular space and let $U$ be an open set such that $A\subseteq U$. Prove that there is an open set $V$ such that $A \subseteq V \subseteq \overline{V} \subseteq U$.

Let $p \in A$ which implies $p \in U$. Then a result is given in the book (Theorem 5.11): A $T_1$-space $(X, \mathcal T)$ is regular if and only if for each member $p$ of $X$ and each neighbourhood $U$ of $p$, there is a neighbourhood $V$ of $p$ such that $\overline{V}\subseteq U$. So, I got $ V \subseteq \overline{V} \subseteq U$.

But I am unable to prove that $A\subseteq V \subseteq \overline{V}$. I thought that I should let $V\subseteq \overline{V} \subseteq A$ but I am not able to find a contradiction.

Can you please help with that?

Of course there need not be any non-empty open subset of $A$ at all, so the last idea is nonsense.. — Henno Brandsma, Jun 10 '21 at 10:19
Find a $V_p$ open such that $p \in V_p \subseteq \overline{V_p} \subseteq U$ for each $p \in A$. Now apply compactness of $A$ (which you haven't used yet)... — Henno Brandsma, Jun 10 '21 at 10:14
@HennoBrandsma Can you please explain your 1st comment in more detail. I don't feel like looking at other answers as I am so close to solution and their approach is different and I want to Continue with my own solution. — , Aug 02 '21 at 13:08
${V_p\mid p \in A}$ is an open cover of $A$ so finitely many, $V_p, p \in F$, $F \subseteq A$ finite also cover $A$. Now define $V=\bigcup_{p \in F} V_p$.. — Henno Brandsma, Aug 02 '21 at 13:12

How do I show that if $A$ is compact and $U \supseteq A$ is open, then there is an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq U$?

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