Here’s a counterexample to the conjecture.
Let $Y=\Bbb N\times\Bbb N$, let $p$ and $q$ be distinct points not in $Y$, and let $X=Y\cup\{p,q\}$. Points in $Y$ are isolated. For each $k\in\Bbb N$ the set
$$B_p(k)=\{p\}\cup\left\{\langle m,n\rangle\in Y:n\ge k\right\}$$
is a basic open nbhd of $p$. For each $k\in\Bbb N$ the set
$$B_q(k)=\{q\}\cup\left\{\langle m,n\rangle\in Y:m\ge k\right\}$$
is a basic open nbhd of $q$. For convenience, for $n\in\Bbb N$ let $S_n=\{n\}\times\Bbb N$.
Each open nbhd of $p$ contains all but finitely many points of each $S_n$, and each open nbhd of $q$ contains all but finitely many of the sets $S_n$. Thus, every open set containing both $p$ and $q$ contains all but finitely many points of $Y$, and $X$ is compact. $Y$ is Hausdorff, as are the sets $B_p(k)$ and $B_q(k)$, so $X$ is locally Hausdorff. However, none of the sets $B_p(k)$ contains a compact nbhd of $p$, so $X$ is not locally compact. (Similarly, $X$ fails to be locally compact at $q$.)
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
Of course, if $X$ is metrizable then it is also locally metrizable.
For the other direction, you could really follow the hint. Supposed that $X$ is locally metrizable, there exists an open cover $\bigcup_{x\in X}U_x$ of $X$ where $U_x$ is a metrizable neighborhood of $x$. Because of compactness, it has a finite subcover: $\bigcup_{i<n} U_i=X$ where $U_i:=U_{x_i}$ for some $x_i$.
Because of metrizability, each $U_i$ has countable base $(V_{i,j})_j$ of open subsets, so that all finite intersections of these are still countable, and they give a base of $X$.
The space also has to satisfy the separation axiom $T_3$, but this holds as $X$ is Hausdorff and compact.