I'm reviewing some topology and have found myself at a bit of a loss after trying to prove the proposition below. Proposition. A space is compact iff every net has an accumulation point.
My problem is that I have produced a "proof" of the incorrect claim: "every cover has a finite subcover if and only if every net has an accumulation point." Of course this is not true since we can always cover a space by its singletons, and not every sequentially compact space is finite. Where is the necessary topological input that I have lost?
"Proof" of $\implies$:
Assume every cover has a finite subcover and towards a contradiction suppose $(x_\alpha)$ has no accumulation point. Then for every $x\in X$ there's a neighborhood not frequented. These neighborhoods yield a cover $(U_x)$, which has a finite subcover $(U_{x_i})$. That $U_{x_i}$ is not frequented means there's a $\beta_{x_i}$ in the directed set beyond which the net does not intersect $U_{x_i}$. By directedness, there's a $\beta$ greater than all the $\beta_{x_i}$, thus $(x_\alpha)_{\alpha\geq \beta}\cap X=\emptyset$. But the net is assumed in $X$ which is a contradiction.
"Proof" of $\impliedby$:
Assume every net has an accumulation point. Let $(B_i)$ be a family of subsets of $X$. To add direction let $(C_\alpha)$ be the family obtained by adjoining their finite intersections ordered by containment. This induces a direction on $A$. For each $C_\alpha$ choose an element $x_\alpha$ and define a net $\alpha\mapsto x_\alpha$. By assumption this net has an accumulation point $x$. Noting $(B_i)$ has the finite intersection property iff $(C_\alpha)$ does, it suffices to assume the latter and prove $\bigcap_\alpha C_\alpha\neq \emptyset$. We prove by contradiction the accumulation point $x$ is in this intersection. If not, it's in its complement and by de Morgan in some $C^c_\alpha$. By directedness, $x_\beta \in C^c_\alpha$ for some $\beta>\alpha$. Thus $x_\beta\in C^c_\alpha\cap V_\beta=\emptyset$ which is a contradiction.
What am I completely missing? Where is openness/closedness needed?
The forward direction is fine. The problem with the other direction is the one noted by Daniel Fischer in the comments: you can’t prove that $x\in\bigcap_\alpha C_\alpha$ unless the sets $C_\alpha$ are closed.
We have a family $\mathscr{C}=\{C_\alpha:\alpha\in\Lambda\}$ of non-empty sets that is closed under finite intersections, so that $\langle\mathscr{C},\supseteq\rangle$ is a directed set, and we have a net $\langle x_\alpha:\alpha\in\Lambda\rangle$ such that $x_\alpha\in C_\alpha$ for each $\alpha\in\Lambda$. Finally, we have an accumulation point $x$ of this net.
Fix $\alpha\in\Lambda$. If $U$ is an open nbhd of $x$, there is a $\beta\in\Lambda$ such that $C_\beta\subseteq C_\alpha$ and $x_\beta\in U$. Evidently $x_\beta\in C_\alpha$, so $U\cap C_\alpha\ne\varnothing$, and it follows that $x\in\operatorname{cl}C_\alpha$. Since $\alpha$ was arbitrary, $x\in\bigcap_{\alpha\in\Lambda}\operatorname{cl}C_\alpha$. If the sets $C_\alpha$ are closed, we can conclude that $x\in\bigcap_{\alpha\in\Lambda}C_\alpha$ and hence that $X$ is compact.
If the sets $C_\alpha$ are not closed, however, $x$ need not be in any of them, and their intersection may indeed be empty even if $X$ is compact, as the following example shows.
Let $X=[0,1]$ with the usual topology, and let $C_n=(0,2^{-n})$ for $n\in\Bbb N$. The family $\mathscr{C}=\{C_n:n\in\Bbb N\}$ is closed under finite intersections and has the finite intersection property. Following the proof, let $x_n\in C_n$ for each $n\in\Bbb N$. The resulting net is actually just the sequence $\langle x_n:n\in\Bbb N\rangle$, which clearly converges to $0$ and therefore has $0$ as an accumulation point; indeed, $0$ is its only accumulation point. However,
$$\bigcap_{n\in\Bbb N}C_n=\bigcap_{n\in\Bbb N}(0,2^{-n})=\varnothing\;.$$