Without assuming that $X$ is Hausdorff (and using the definition of local compactness given) the result is not true.
Let $X$ be any set with at least two elements, and consider the trivial (anti-discrete) topology on $X$. Clearly this space has the property that the nonempty open sets with compact closures form a base (indeed, there is only one nonempty open set, and it's closure is clearly compact), and so this space is locally compact. Note, too, that any nonempty proper $M \subseteq X$ is also locally compact (another anti-discrete space), but will not be the intersection of an open and a closed subset of $X$.
However, the result does follow if we assume that $X$ is Hausdorff. (Then local compactness is equivalent to every point having an open neighbourhood with compact closure.)
To wit: given $x \in M$, let $V_x$ be a neighbourhood of $x$ in $M$ such that $\mathrm{cl}_M ( V_x ) = \overline{V_x} \cap M$ is compact. Now $\overline{V_x} \cap M$ is also a compact subset of $X$, so by Hausdorffness $\overline{V_x} \cap M$ is closed (in $X$). Fix an open $W_x \subseteq X$ such that $W_x \cap M = V_x$. Clearly we have that $$\overline{ W_x \cap M } \cap M = \mathrm{cl}_M (V_x).$$ Note, now, that as $W_x \cap M \subseteq \overline{ W_x \cap M } \cap M$, we have that
$$
W_x \cap \overline{M}
\subseteq \overline{ W_x \cap \overline{M} }
= \overline{ W_x \cap M }
\subseteq \overline{ W_x \cap M } \cap M
\subseteq M,
$$
(and clearly $x \in W_x \cap \overline{M}$).
Setting $U = \bigcup_{x \in M} W_x$, it follows that $U$ is open in $X$, and $M = U \cap \overline{M}$, as desired.
Looks fine, for the most part, though I have no idea what you could mean by $U\cap A$ if $U$ is an open cover of $A$. I suspect that you instead mean that $C$ is a compact subset of $X$ containing an open neighborhood $U$ of $x.$ Also, closed subsets of Hausdorff spaces need not be compact (consider $\Bbb R$ as a subset of itself, for example), though compact subsets of Hausdorff spaces will be closed.
Alternately, you can use your previous result to proceed even more directly in showing that $C\cap A$ is compact. Since $A$ is closed, then $C\cap (X\setminus A)$ is open in $C,$ so $C\cap A=C\setminus \bigl(C\cap(X\setminus A)\bigr)$ is closed in $C,$ so is compact by your previous result since $C$ is compact.
Best Answer
You want to show that if $K$ is a closed subset of $X$, $x\in K$, and $U$ is a compact nbhd of $x$ in $X$, then $U\cap K$ is a compact nbhd of $x$ in $K$. You’ve done everything except show that $U\cap K$ is compact.
To show this, let $\mathscr{V}$ be an open cover of $U\cap K$; you want to find a finite subcover. $K$ is closed in $X$, so $X\setminus K$ is open; let $\mathscr{W}=\mathscr{V}\cup\{X\setminus K\}$. $\mathscr{W}$ is a collection of open sets.
Then you’ll know that some finite $\mathscr{W}_0\subseteq\mathscr{W}$ covers $U$, since $U$ is compact. Clearly $\mathscr{W}_0$ covers $U\cap K$. If $X\setminus K\notin\mathscr{W}_0$, then $\mathscr{W}_0\subseteq\mathscr{V}$, and you have the subcover that you wanted.