A property about locally compact Hausdorff spaces

Question

A space $X$ is said to be locally compact if for every point $x$ of $X$ there exists an open set $U$ and a compact set $K$, such that $x\in U\subseteq K$.

What I want to show is that (which I think is true)

If $X$ is a locally compact Hausdorff space, then for every point $x$ of $X$ and every compact set $K$ containing $x$ there exists an open set $U$, such that $x \in U \subseteq K$.

Things which might help:

1. A locally compact Hausdorff space is regular.

2. If $X$ is a locally compact Hausdorff space, then for every point $x$ of $X$ and every open set $U$ containing $x$ there exists a compact set $K$, such that $x \in K \subseteq U$.

Here is the proof of 2)

Let $x$ be a point of $X$ and let $U$ be an open neighbourhood of $x$. Then $X\setminus U$ is closed and $x \notin X\setminus U$. Since $X$ is regular, there exist open sets $U_1$ and $U_2$ such that $x\in U_1$ and $X\setminus U \subseteq U_2$ and $U_1\cap U_2=\emptyset$. Let $K_1=X\setminus U_2$. Then $K_1$ is closed and $x\in U_1\subseteq K_1 \subseteq U$. Since $X$ is locally compact, there exists an open set $V$ and a compact set $K_2$ such that $x \in V \subseteq K_2$. Since $X$ is Hausdorff, we have $K=K_1 \cap K_2$ is a compact set, such that $x\in K \subseteq U$.

Edit: If I can replace compact set with compact neighborhood in 2), then my purpose will suffice.

Edit 2: In proof of 2) $x \in U_1 \cap V \subseteq K \subseteq U$. Hence $K$ is a compact neighborhood of $x$.

Answer 1

Your claim is false. Let's take $\mathbb{R}$ as our topological space for this example. $1 \in [0,1] \subset \mathbb{R}$, but no open (in $\mathbb{R}$) set $U \subset [0,1]$ containing 1.

Hope this helps!

Answer 2

If $x \in K$, $K$ compact, either $x \in \operatorname{int}(K)$ and then trivially $U = \operatorname{int}(K)$ works as your open set, or $x \notin \operatorname{int}(K)$ and then no such $U$ can exist (becuase it would make $x$ an interior point of $K$). So your proprosed property is not really relevant: it either trivially holds or cannot hold anyway (e.g. as in $K=\{x\}$ when $x$ is not an isolated point of $X$).

It has nothing to do with local compactness at all. Local compactness (as you define it) just says that every point is in the interior of at least one compact $K$, nothing more. But in general it won't be an interior point of any compact set it's in, as you wanted!