$X$ is Hausdorff space. The following is equivalent.
(a) $X$ is locally compact space.
(b) For every open neighborhood $U$ of $x\in X$ there is a smaller open neighborhood $V$ of $x$ whose closure is compact and is contained in $U$.
I tried the following to show $(a)\Rightarrow(b)$.
Try :
Since $X$ is a locally compact hausdorff space, there exists an open set $W$ of $X$ where $x$ $\in$ $W$ $\subset$ $\overline{W}$ and $\overline{W}$ are compact for a given point $x \in X$.
Then, since $X$ is a regular space, subspace $\overline{W}$ is also a regular space.
Since $U\cap \overline{W}$ is an open set of $\overline{W}$, there exists an open set $O$ of $\overline{W}$ that satisfies $x \in O \subset \overline{O} \subset U\cap \overline{W}$.
Then there exists an open set $G$ of $X$ with $O=G\cap \overline{W}$.
Now, $V=G\cap W$ is an open set of $X$, and $\overline{V}$ is a closed subset of the compact space $\overline{W}$, so $\overline{V}$ is compact.
Q. How do you prove $\overline{V} \subset U$?
It does not look obvious!
Please help me. No matter how worried, it does not improve.
You use in your proof that the space is regular and hence every neighborhood contains a closed neighborhood. But if you know how to prove this, then you can easily prove your claim without that much trouble:
We may assume without loss of generality that $U$ is contained in $W$ (otherwise intersect it with $W$, then find such a $V$ and this very $V$ will do the job for the original $U$).
Let $V$ be the interior of any closed neighborhood of $x$ inside of $U$. Then, since $\bar{V}$ is inside of $\bar{W}$ compact, $\bar{V}$ is also compact. And since $V$ is the interior of a closed neighborhood contained in $U$, $\bar{V}$ is also contained in $U$.
Edit: regarding your original approach, to see the inclusion $\bar{V}\subset U$, you can check that $\bar{V}\subset \bar{O}$ ($\subset U$). To see this inclusion, let $y\in \bar{V}$, i.e. any neighborhood of $y$ meets $V=G\cap W$. But then it also meets $G\cap \bar{W}=O$. Therefore $y\in \bar{O}$.