Locally compact subspace is an intersection of an open and closed set Let X be a locally compact topological space. I need to prove that if $M\subset X$ is a locally compact subspace of X then there exist $U,F\subset X$ such that U is open and F is closed, and $M=U\cap F$.
It can be assumed that for every open $x\in V$ there exist an open subset U such that $x \in U \subset \overline U \subset V$ and $\overline U$ is compact.
My thoughts about the problem are that for every $x\in M$ and for every neighborhood $x\in V_x$ there exist an open neighborhood $U_x$ in M such that $x\in U_x\subset \overline U_x \subset V_x$ and $\overline U_x$ is compact and closed in M. So there exist an open subset $Z_x\subset X$ such that $U_x = Z_x\cap M$. I can take $U = \bigcup _{x\in M}Z_x $, then U is open in X, but i was having trouble finding the closed set F. I thought  to take $F = \overline {\bigcup _{x\in M}U_x}$, that is closure in X. Now $M\subset U\cap F$, but I haven't succeeded in showing that $M\supset U\cap F$, maybe that's not even true.
Thanks!

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.
