Let $(x_n)$ be a sequence in a metric space X. Show that $x_n$ converges to x in X iff for every open set containing x there is a natural number N so that if $n \geq N$ $x_n \in U$
Suppose $x_n \rightarrow x$. Let U be an arbitrary open set containing X. Hence there exists a radius $r>0$ so that $B(x,r) \subset U$. Since $x_n$ converges to x, there exists a natural number N so that if $n\geq N$ then $x_n \in B(x,r) \subset U$
Suppose for each open set U containing x there is a natural number N so that when $n \geq N$ we have $x_n \in U$ Let $\epsilon >0$ be arbitrary. Let $U=B(x,\epsilon)$. Since U is an open subset of X, there is a natural number N so that whenever $n\geq N$ , $x_n \in B(x,\epsilon)$. Since epsilon is artbitrary, $x_n$ converges to x.
Is my proof correct?
Yes that's correct. In essence, we only need to check convergence of a sequence using basic open sets (and open balls form a base for the metric topology).
(If you know subbases, then note we could have used subbasic sets too, as we can use finite maxima of indices to talk sequences into induced base elements..)