I'm currently working through "Tensor Analysis on Manifolds" by Richard L. Bishop and Samuel I. Goldberg and came across proposition 0.2.8.2 (a) which I uploaded here. It's the proof's last sentence that caught my eye. On the one hand it appeared intuitive that every neighborhood of a point x of a subset's closure intersects the subset. On the other hand I wanted proof.
So I went back a few pages to look whether I had missed something and, indeed, I had: On this page, the authors give an alternative definition of closed sets as well as the closure and the boundary of a set in case a basis of neighborhoods of the topological space X is given. Still, as all of this belongs to a short recap of topology, they gave no proof.
…which is why I tried to proof it myself. First of all, Bishop and Goldberg defined open and closed sets as follows: Open sets are given by the topology T, closed ones are the respective complements of the first. The interior and the closure of a subset A are:
$\begin{align} A^0 &= \bigcup \text{All open sets contained in}~ A \\ \overline{A} &= \bigcap \text{All closed sets containing}~ A \end{align}$
While I had no trouble showing the statement about closed sets, my proof of the equivalence of above's definition for $\bar{A}$ and the alternative one ("The closure of A consists of those x such that every basis neighborhood of x intersects A") went like this:
"$\subset$": Let $x \in A^0$ (as defined above) and let U be a basis neighborhood. Show that $U \cap A \neq \emptyset$.
Assume $U \cap A = \emptyset \Rightarrow U \subset X-A$. W.l.o.g let U be open (otherwise take $U^0$ instead) $\Rightarrow X-U$ is closed. Also, since $U \cap A = \emptyset$, it follows that $A \subset X-U$, which contradicts the fact that $x$ is in every closed superset of A.
The part I'm anything but sure about is the "w.l.o.g let $U$ be open". Again, it appeared intuitive in the first moment but, thinking about it again and considering that the topology is induced by the basis of neighborhoods (which, as I understand, doesn't necessarily consist of open neighborhoods), I couldn't come up with an explanation that $U^0 \neq \emptyset$ (let alone $x \in U^0$) if I don't assume that $X$ is Hausdorff (like in proposition 0.2.8.2 (a)).
But maybe I'm on the wrong path? Anyway, I'd be glad if someone of you could shed some light on that matter.