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> Let $S\subset M$ dense subset. Given a sequence $(x\_n)$ in $M$, suppose that for some $x\in M$ $\lim d(x\_n,s) = d(x,s)$ for every $s\in S$. Prove that $\lim x\_n = x$.
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My attempt: Since $S$ is dense in $M$ and $x\in M$, it follows that $x = \lim a\_n$ for some sequence $(a\_n)$ in $S$. It is given that $\lim d(x\_n,s) = d(x,s)$ for every $s\in S$. So, in particular we can evaluate for all elements of $(a\_n)$. Therefore, noting that the metric function $d$ is continuous, we obtain $\lim d(x\_n,a\_n) = d(x,a\_n) = d(\lim a\_n,a\_n) = \lim d(a\_n,a\_n) = 0.$ 

But this only proves that $\lim d(x\_n,a\_n) = 0$. How can I proceed to the conclusion that $\lim x\_n = x$?