Proof based on distance function continuity I've proved the continuity of the distance function $d:X \times X\rightarrow \mathbb{R}$ in metric space $(X, d)$. Now I've to work on this: 

Let $S \subseteq X$ be a dense set and $\{x_n\}$ a sequence included in $X$. Prove that if there exists $x \in X$ such that $\lim\limits_{n \to \infty} d(x_n, s) = d(x, s)$ for all $s \in S$, then $\lim \limits_{n \to \infty} x_n = x$.

My proof: We want to prove that $\lim \limits_{n \to \infty} d(x_n, x) = 0$. Since $S$ is dense, let $\{s_m \}$ be a sequence in $S$ such that $s_m \to x$. 
$$
\begin{align}
\lim_{n \to \infty} d(x_n, x) 
&= \lim_{n \to \infty} d \left(x_n, \lim_{m \to \infty}  s_m \right) \\ 
&= \lim_{n \to \infty} \ \lim_{m \to \infty}  d(x_n, s_m) \\ 
&= \lim_{m \to \infty} \ \lim_{n \to \infty} \ d(x_n, s_m) & & (d \text{ is continuous)} \\ 
&= \lim_{m \to \infty} \ d(x, s_m) && \text{(by hypothesis)}\\ 
&= d(x, x) = 0.
\end{align}
$$
Do you think this proof is correct? It is heavily bases in the fact that $d$ is continuous.
I'll really appreciate any advice.
Thanks!
 A: Two remarks.
1) Your sequence $(s_m)$ is not a sequence. Because $S$ is dense you can pick some sequence such that $s_m \to x$.
2) You should explain why you can change order of limits.

However, proof can go like this.
Since $S$ is dense, for every $\epsilon > 0$ you can pick $s \in S$ such that $d(s,x) < \epsilon$. Then
$$
d(x_n,x) \leq d(x_n,s) + d(s,x) \leq \epsilon + \epsilon = 2\epsilon
$$
for sufficiently large $n$ by assumption. Hence $x_n \to x$.
A: I think you don't need the continuity of $d$. You need to show that $d(x_n,x)\to 0$. So, let $\epsilon\gt 0$. Since $S$ is dense in $X$, exist $s\in S$ such that $d(s,x)\lt \epsilon/4$. Since $d(x_n,s)\to d(x,s)$, there exist a $N\in\mathbb{N}$ such that if $n\geq N$ then $|d(x_n,s)-d(x,s)|\lt \epsilon/2$. Now, if $n\geq N$
$$\begin{align*}
d(x_n,x) &\leq d(x_n,s)+d(s,x)\\
&= |d(x_n,s)-d(x,s)+d(x,s)|+d(s,x)\\
&\leq |d(x_n,s)-d(x,s)|+|d(x,s)|+d(s,x)\\
&= |d(x_n,s)-d(x,s)|+ 2d(s,x)\\
&\lt \frac{\epsilon}{2}+2\cdot\frac{\epsilon}{4}=\epsilon.
\end{align*}$$
Since $\epsilon$  is arbitrary, this proves that $d(x_n,x)\to 0$.
