Let $(X,d)$ be a metric space and let $E \subset X$.
Prove: $x$ is an adherent point of $E$ if every open neighborhood of $x$ has a non-empty intersection with $E$.
What I have so far:
Define the open ball $B(x,\epsilon) := \{ y: d(x,y) < \epsilon\}$
If $x \in E$, we are done (as every point in $E$ is an adherent point of $E$). So assume $x \notin E$.
Fix $\epsilon > 0$. Suppose $y \in B(x,\epsilon) \cap E$.
Since $B(x,\epsilon)$ is open, and $y \in B(x,\epsilon)$, there exists $\epsilon_1 > 0$ such that $B(y,\epsilon_1) \in B(x,\epsilon)$.
At this point, if we can establish that some points in $B(y, \epsilon_1)$ are also in $E$, we can say there is a sequence of points $\{x_n\}_{n \in \mathbb{N}} \subset E$ that is within $\epsilon$ of $x$, so that $\lim_{n \to \infty}x_n = x$. Then $x$ is a limit point of $E$, and hence is an adherent point. But how can we say that there are points in $B(y,\epsilon_1)$ that are also in $E$?
Thanks.
Take the open balls of radius $1/n$ centered at $x$. For all $n$ by hypothesis we know that exists $x_n \in E \cap B_{1/n}(x)$. Then the sequence $\{ x_n \} \subset E$ converges to $x$.
EDIT: I corrected the proof.