I was reading the book "Rings of Continuous Functions" by Leonard Gillman when I saw the next exercise. I can't understand the proof in the book. Can someone explain to me? I really appreciate any help you can provide.
Section 3.15. Locally compact spaces. Page 45.
Let $X$ be a subspace of a Hausdorff space $T$.
$(b)$ If $X$ is dense in $T$, then every compact neighborhood in $X$ of a point $p\in X$ is a neighborhood in $T$ of $p$.
Proof
Let $U$ be the interior of a compact neighborhood of $p$ in $X$. Then $\text{cl}_{X}(U)$ is compact, hence closed in $T$, so that, $\text{cl}_{T}(U)=\text{cl}_{X}(U)$. Let $V$ be an open set in $T$ such that $V\cap X=U$. Since $X$ is dense, we have $\text{cl}_{T}(V)=\text{cl}_{T}(U)\subset X$, so that $V=U$.
First, I understand that if $W$ is the compact neighborhood of $p$ in $X$ and we take $U=\text{int}_{X}(W)$ then, $\text{cl}_{X}(U)=\text{cl}_{X}(\text{int}_{X}(W))\subseteq \text{cl}_{X}(W)=W$ The last equality is because $W$ is a compact set in a $T_2$ space. We can conclude that $\text{cl}_{X}(U)$ is a compact set because is a closed subset of a compact set. Moreover, this implies that $\text{cl}_{X}(U)$ is closed in $T$, so that, $\text{cl}_{T}(U)=\text{cl}_{X}(U)$. By definition, because $U$ is open in $X$, there exist $V$ an open set in $T$ such that $V\cap X=U$. This is the part where I am so stuck. How can we conclude that if $X$ is dense then $\text{cl}_{T}(V)=\text{cl}_{T}(U)$? Moreover, I can't see how can we derive that $V=U$.
Since $U\subset V$, $\operatorname{cl}_T(U)\subset\operatorname{cl}_T(V)$.
Now, let $y\in\operatorname{cl}_T(V)$ and let $W$ be an open subset of $T$ such that $y\in W$. Since $V\cap W$ is an open set and $X$ is dense in $T$, there's some $x\in X$ such that $x\in V\cap W$. And, since $V\cap X=U$, $x\in U$. So, this proves that any open subset of $T$ containing $y$ contains an element of $U$. Therefore, $y\in\operatorname{cl}_T(U)$.