Full question:
$\left< M,\rho \right>$ is a compact metric space with sequences $\{x_n\}_{n=1}^\infty$, $\{y_n\}_{n=1}^\infty$ and $\displaystyle{\lim_{n \to \infty} \rho(x_n, y_n)=0}$.
Prove $\exists$ subsequences $\{x_{n_k}\}_{k=1}^\infty$, $\{y_{n_k}\}_{k=1}^\infty$ and a point $x \in M$ s.t.
$\displaystyle{\lim_{k \to \infty} \rho(x_{n_k}, x)=0}$ and $\displaystyle{\lim_{k \to \infty} \rho(y_{n_k}, x)=0}$
Below is my attempt and I would like to know if there is any mistake or room for improvement.
Proof:
$M$ is compact so $\{x_n\}_{n=1}^\infty$ has a subsequence $\{x_{n_k}\}_{k=1}^\infty$ that converges to a point in $M$. Call this point $x \in M$.
So $\displaystyle{\lim_{k \to \infty} x_{n_k}=x}$ $\implies$ $\displaystyle{\lim_{k \to \infty} \rho(x_{n_k}, x)=0}$
So given an $\epsilon \gt 0$, $\exists N_1 \in \mathbb{N}$ s.t. if $k \geq N_1$ then $\rho(x_{n_k}, x) \lt \frac{\epsilon}{2}$
Also given $\displaystyle{\lim_{n \to \infty} \rho(x_n,y_n)=0}$, so $\exists N_2 \in \mathbb{N}$ s.t. if $n \geq N_2$ then $\rho(x_n,y_n) \lt \frac{\epsilon}{2}$
But $n_k \geq n \ \ \forall n,k \in \mathbb{N}$
So $\rho(x_{n_k},y_{n_k}) \lt \frac{\epsilon}{2}$ if $k \geq max\{N_1,N_2\}$
Now, let $k \geq max\{N_1,N_2\}$ then
$\rho(y_{n_k}, x) \leq \rho(y_{n_k},x_{n_k}) + \rho(x_{n_k}, x)$ $\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
So for any given $\epsilon \gt 0$ we have shown there is an index $N$ s.t. if $n \geq N $ then $\rho(y_{n_k}, x) \lt \epsilon$ and $\rho(x_{n_k}, x) \lt \epsilon$ which means
$\displaystyle{\lim_{n \to \infty} \rho(x_n,x)=0}$ and $\displaystyle{\lim_{n \to \infty} \rho(y_n,x)=0}$ $\ \ \ \ \ \ \Box$