Let $M$ be a metric space and let $p\in M$ and $\{x_n\}$ a sequence in $M$ so $x_n\to p$ iff for all neighborhood $u$ of $p$ there is $n_0$ such that for all $n_0<n_1: x_{n_1}\in u$
$\Rightarrow:$ Let there be an open neighborhood $u$ of $p$ be definition of an open set around each element of the set there in an open ball such that the ball is contained in the set, in particular $p\in u$ so there is an open ball $B(p,\epsilon)$ such that $B(p,\epsilon)\subseteq u$ now because $x_n\to p$ for all $\epsilon>0$ there is $n_0$ such that for all $n_0<n_1:x_{n_1}\in B(p,\epsilon)$ so for all $u$ we can find $\epsilon$ s.t for all $n_0<n: x_n\in u$
$\Leftarrow:$ For all neighborhood $u$ of $p$ there is $k_0$ such that for all $k_0<k_1: x_{k_1}\in u$ let $u$ be the open ball around $p$ namely $B(p,\epsilon)$ so $x_{k_1}\in B(p,\epsilon)$ and by definition $x_n\to p$
Is the proof redundant all that that we have to say is that an open neighborhood $u$ is an open ball and vice versa?