Showing $x\in \overline{A}$ if and only if every open set $U$ containing $x$ intersects $A$ 'directly'? Let $A$ be a subset of a topological space $X$. It can be shown that $x\in \overline{A}$ if and only if every open set $U$ containing $x$ intersects $A$.
I can prove this by proving the contrapositive:
$x\notin \overline{A}$ if and only if there exists an open set $U$ containing $x$ that does not intersect $A$.
But is this the only way to prove this theorem? That is, can it only be shown using the contrapositive statement, or is there a more 'direct' way to prove it?
 A: Defining $\overline{A}  =\bigcap\{C \subseteq X \text {closed}: A \subseteq C\}$
(the smallest closed set containing $A$ as a subset):
Let $x \in \overline{A}$ and let $U$ be an open neighbourhood of $x$. If $U \cap A = \emptyset$ then $A \subseteq X\setminus U$, and the latter set is closed, so $\overline{A} \subseteq X\setminus U$, in this definition. Contradiction, as witnessed by $x$. So $A \cap U \neq \emptyset$.
Suppose $x$ has the neighbourhood intersection property. Let $C$ be a closed set that contains $A$. If $x \notin C$, then $X \setminus C$ is a neighbourhood of $x$ that misses $A$, which cannot be. So $x \in C$, and as $C$ was arbitrary closed around $A$,  $x \in \overline{A}$.
Pretty direct I'd say. We need small proofs by contradiction, but this is inherent to the statement, I think. To show $U \cap A \neq \emptyset$, we either have to find a concrete point that witnesses it or assume the opposit to make progress.
A: Let's take the definition of $\overline{A}$ to be the union of $A$ with the limit points of $A$. We can write a very clean direct proof using nets. One direction is easy: if $x\in \overline{A}$, then any open set containing $x$ has a nonempty intersection with $A$. If $x\in A$, this is obvious, so let $x$ be a limit point of $A$. Then, there is some net $\langle x_{\alpha}\rangle_{\alpha\in \mathcal{A}}\subseteq A$ such that for any neighborhood $U$ of $x$, there is an $\alpha\in \mathcal{A}$ such that $\beta\geq \alpha$ implies $x_{\beta}\in U$. Clearly, this implies that for this given $U$, $x_{\alpha}\in U$ (since $\alpha\geq \alpha$), so $U\cap A$ must be nonempty.
Now, if every open set containing $x$ has a nonempty intersection with $A$, then for the family $\{U_{\alpha}\}_{\alpha\in \mathcal{A}}$ of neighborhoods of $x$ with $U_{\beta}\subseteq U_{\alpha}$ iff $\beta\geq \alpha$, we consider a net $\langle x_{\alpha}\rangle_{\alpha\in\mathcal{A}}$ such that $x_{\alpha}\in U_{\alpha}\cap A$. Then, for any neighborhood $U_{\alpha}$ of $x$, $x_{\beta}\in U_{\alpha}$ for all $\beta\geq \alpha$, so $x_{\alpha}\to x$. Thus, $x$ is a limit point of $A$, so $x\in \overline{A}$.
