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I am trying to prove the following:

If $E \subset X$, then $x \in \overline{E}$ iff there is a filter base in $E$ converging to $x$.

I can only prove the converse implication, as follows. (Any feedback on this would be great).

Conversely, suppose there is a filter base $\mathcal{B}$ in $E$ converging to $x$. We have to show that $x$ is a point of closure of $E$. Let $U(x)$ be an open neighbourhood of $x$. Since $\mathcal{B}$ converges to $x$, there exists $B' \in \mathcal{B}$ such that $b \in B' \subset U(x)$. It follows that $b \in U(x)\cap E$. Thus, $U(x) \cap E \neq \emptyset$ for every open neighbourhood $U(x)$ of $x$. Thus, $x$ is a point of the closure of $E$, and hence $x \in \overline{E}$.

How do I prove the direct implication, i.e., if $x \in \overline{E}$, then there is a filter base in $E$ converging to $x$?

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For the direct implication one can show that if $x \in \overline{E}$, then $$\mathcal{B} = \{ U \cap E : U \text{ is an open neighbourhood of } x \}$$ is a filter base in $E$ converging to $x$.

Your proof of the reverse implication seems fine to me, although I would rephrase part of it thusly:

... Since $\mathcal B$ converges to $x$, there exists a $B^\prime \in \mathcal B$ such that $B^\prime \subseteq U(x)$, and so in particular $\emptyset \neq B^\prime \subseteq U(x) \cap E$. ....

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