Compare definitions of locally compact spaces

For some point $x$ take its compact neighborhood $K$. We want to show that for any open neighborhood $U$ of $x$, there is some compact neighborhood $K'\subseteq U$ and we will use $K$ to construct it.

So, take any open neighborhood $U$ of $x$. If $K\subseteq U$, we are done. Otherwise, $K\cap U^c$ is closed subset of $K$, thus compact. Since space is Hausdorff, we can find open $V$ and $V'$ such that $K\cap U^c\subseteq V$ and $x\in V'$, $V\cap V'=\emptyset$. Define $K' = K\cap V^c \subseteq K$. You can easily check that $K'$ is compact neighborhood of $x$ contained in $U$.

Edit:

Let me clarify, in Hausdorff space you can separate compact sets from points. Let $K$ be compact and $x\not\in K$. For each $y\in K$ choose open $y\in U_y$, $x\in V_y$ such that $U_y\cap V_y=\emptyset$. $\{U_y\mid y\in K\}$ is open cover for $K$, so we can find finite subcover $\{U_{y_1},\ldots,U_{y_n}\}$ of $K$. Define $U=\cup_{i=1}^n U_{y_i}$, $V=\cap_{i=1}^n V_{y_i}$. Then $K\subseteq U$, $x\in V$, $U\cap V =\emptyset$.