Exercise: Let $X$ be a topological space such that every point has a compact neighbourhood. Prove that the set of compact subspaces of $X$ is a fundamental cover of $X$.
Proof: Let $E \subset X: E \cap K$ is open in $K \ \forall$ compact subspace $K$ of $X$, and for all $x \in X$ let $K_x$ be a compact neighbourhood of $x$. Then, $\exists A_x \subset X, V_x \subset K_x$ open sets such that $E \cap K_x = A_x \cap K_x$ and $x \in V_x$. We now have $$E= \bigcup_{x \in E}\{ E \cap K_x \} = \bigcup_{x \in E}\{ A_x \cap K_x \} \supseteq \bigcup_{x \in E}\{ A_x \cap V_x \} .$$ Since $x \in A_x \cap V_x \forall x \in E$, we have $E \subseteq\bigcup_{x \in E}\{ A_x \cap V_x \}$, so $E=\bigcup_{x \in E}\{ A_x \cap V_x \}$ which is the union of open sets and is therefore open.
I never used the fact that $K_x$ is compact though, so where is my mistake?
You did use the fact that $K_x$ is compact when you implicitly claimed that $E ∩ K_x$ is open in $K_x$. On the other hand, this was necessary only because you wanted to prove that the set of all compact subspaces is a fundamental cover. In fact, instead of the set of compact subspaces you may use any family $\mathcal{A}$ of sets, and the proof works the same, and gives you that $\mathcal{A}$ is a fundamental cover. Note that this is quite easy – every open cover is fundamental, and every cover that is refined by a fundamental cover is itself fundamental. And each point has a neighborhood in a cover $\mathcal{A}$ if and only if $\mathcal{A}$ is refined by an open cover.