- $(x_n)$ converges to $a$ if for every $\varepsilon >0$ there is a $N$ such that $n\ge N \Rightarrow \text{d}(x_n, a) <\varepsilon,$ where $\text{d}$ is the metric function associated to the metric space in question.
- By "$B(a,\varepsilon)$" I mean the ball of radius $\varepsilon$ centered at $a$, which is simply $\lbrace x \text{ in your metric space}: \text{d}(x,a)<\varepsilon \rbrace$
$(\Rightarrow)$
Suppose $(x_n)$ converges to $a$. Then for every $\varepsilon>0,$ there is a $N$ for which $n\ge N \Rightarrow \text{d}(x_n, a)<\varepsilon.$ Let $U$ be any open neighborhood containing $a$. Since $U$ is open, there is an $\varepsilon >0$ for which $B(a,\varepsilon)\subset U$. Therefore, $n\ge N$ impiles $x_n\in B(a,\varepsilon)\subset U$.
$(\Leftarrow)$
By hypothesis, every open set $B(a, \frac{1}{n})$ contains the tail of some sequence $(x_n)$. Let $\varepsilon >0$. There is an $N$ for which $n\ge N \Rightarrow \frac{1}{n}<\varepsilon.$ Therefore, $n\ge N$ implies that $\text{d}(x_n,a)\le \frac{1}{n}<\varepsilon$, since $x_n \in B(a, \frac{1}{n})$ for $n\ge N.$ Therefore, $x_n \rightarrow a$.