$F:X \to Y$ open continuous map when $X$ is Hausdorff and locally compact and $Y$ is Haussdorf, show that $F(X)$ is locally compact.
I took some $F(x)$ for $ x\in X $ and open neighbourhood $ W \subset Y , F(x) \in W$ because of the continuity of $F$ we have $F^{-1}(W)$ as a neighbourhood of x, because $X$ is locally compact Hausdorff I have open neighbourhood of x: $U\subset X$ such that $x \in U \subset \overline{U} \subset F^{-1}[W]$.
Does the map preserve containment under open map? I know that $F[\overline{U}]$ is compact and that $F(\overline{U}) \subset \overline{F[U]}$
Does $ F(x) \in F[\overline{U}] \subset \overline{F[U]} \subset W$ and it's enough?
Given the definition, your argument is about correct: let $W$ be an open neighbourhood of $F(x) \in F[X]$. Then $x \in F^{-1}[W]$ and the latter set is open by continuity of $F$, so local compactness of $X$ gives an open neighbourhood $U$ of $x$ with $$x \in U \subseteq \overline{U} \subseteq F^{-1}[W]$$ and $\overline{U}$ compact.
But then $F[\overline{U}]$ is compact by continuity of $F$ and as images preserve inclusions:
$$F(x) \in F[U] \subseteq F[\overline{U}] \subseteq \overline{F[U]}$$
but we only know that $F[\overline{U}] \subseteq W$, but not yet for the last set. To solve that, apply the local compactness in $X$ again for $x$ and $U$ and use the $V$ we get there: $F[V]$ is as required, as $\overline{F[V]} \subseteq F[U] \subseteq W$ which is enough. I don't see that we need Hausdorffness for $Y$ or Hausdorffness of $X$ for this proof. I suppose they're needed for local compactness.