Hausdorff space in which every point has a compact neighbourhood is compactly generated A topological space $X$ is said to be compactly generated if, defined $\mathcal{K} : = \{ K \subset X | \text{$K$ is compact} \}$, $X= \bigcup\{K | K \in \mathcal{K} \}$ and $U \subset X$ is open iff $U \cap K$ is open in $K$ for every $K \in \mathcal{K}$.
Prop: If $X$ is a topological Hausdorff space in which every point has a compact neighbourhood then $X$ is compactly generated.
Proof (must be wrong, I do not use the Hausdorff hypothesis):
Let $x \in X$, $\mathcal{I}(x) := \{ V \subset X | \text{$V$ is a neighbourhood of $x$} \}$. Let $K_x$ be a compact neighbourhood of $x$: there exist $V_x$ open in $X$ s.t. $x \in V_x \subset K_x$.The family $\mathcal{A}_x := \{ A \subset X : A = V \cap V_x | V \in \mathcal{I}(x) \}$ is a family of open neighbourhoods of $x$. For every $V \in \mathcal{I}(x)$ there is $A \in \mathcal{A}$ s.t. $A \subset V$.
Let be $U \subset X$ s.t. $U \cap K$ is open in $K$ for every $K$ compact subset of $X$. Let $x \in U$ and consider $K_x$ compact neighbourhood of $x$. Then $U\cap K_x$ is open in $K_x$ that is there exists $V$ open in $X$ s.t. $U \cap K_x = V \cap K_x$. Now, $x \in U \cap K_x = V \cap K_x$ which is a neighbourhood of $x$. There there is $A \in \mathcal{A}$ s.t. $x \in A \subset V \cap K_x$. And then $V$ is an open set containing $x$ and $V \subset U$. Then $U$ is open.
 A: I think your proof is correct. I would go a different route in the proof:
Suppose that $F$ is such that $F \cap K$ is closed in $K$ for any compact subset $K$ of $X$. It suffices to  show $F$ is closed in $X$.
Suppose not, then there is some $p \in \overline{F}\setminus F$. This $p$ has a compact neighbourhood $K_p$, so $K_p$ compact and for some open set $O_p$ we have $p \in O_p \subseteq K_p$. By assumption $F \cap K_p$ is closed in $K_p$. But $p \in \overline{(F \cap K_p)}$ (in $K_p$) while $p \notin F \cap K_p$. (To see the first claim: if $U \cap K_p$ is an open neighbourhood of $p$ in $K_p$ with $U \ni p$ open in $X$, then $U \cap O_p$ is open in $X$, contains $p$, so intersects $F$,  and as $O_p \subseteq K_p$, it intersects $F \cap K_p$.) 
This contradiction shows that $F$ is indeed closed in $X$.
Or following your idea, in a different (I think clearer) writeup : if $O$ is any subset with $O \cap K$ open in $K$ for any compact subset $K$. Then pick $p \in O$, and a compact neighbourhood $K_p$ of $p$ (so we have $p \in O_p \subset K_p$ with $O_p$ open in $X$). Also, $O \cap K_p$ is open in $K_p$ by assumption, so we have $O'$ open in $X$ such that $O' \cap K_p = O \cap K_p$. Then $$p \in O' \cap O_p \subseteq O' \cap K_p = O \cap K_p \subseteq O$$
As $O' \cap O$ is open in $X$, this shows that $p$ is an $X$-interior point of $O$, and as $p \in O$ is arbitary, $O$ is open in $X$.
So when we use the definition of locally compact as "every point has a compact neighbourhood" (neighbourhood in the most general sense), we don't need Hausdorffness. But we do need Hausdorffness (or at least some separation axiom) to see that this particular definition is equivalent to "every point has an open neighbourhood with compact closure", or "any point has a local base of open neighbourhoods with compact closure", all of which are commony used as well. The Hausdorffness makes those (subtle) differences moot.
