Let $x_{n}$ and $y_{n}$ be two sequences in a metric space $(X,d)$. Suppose that $x_{n}$ converges to a point $x\in X$ and $y_{n}$ converges to a point $y\in X$. Show that $\displaystyle\lim_{n\rightarrow\infty}d(x_{n},y_{n}) = d(x,y)$.
MY ATTEMPT
Lemma
Given any three points $x,y,z\in X$, one has that $d(x,y) \geq |d(y,z) - d(x,z)|$.
Proof
According to the triangle inequality, one has that $d(x,y) + d(y,z) \geq d(x,z)$.
Similarly, $d(z,x) + d(x,y) \geq d(z,y)$. Thus
\begin{align*} \begin{cases} d(x,y) \geq d(x,z) - d(y,z)\\\\ d(x,y) \geq d(y,z) - d(x,z) \end{cases} \Rightarrow d(x,y) \geq |d(x,z) - d(y,z)| \end{align*}
My solution
According to the definition of convergence, for every $\varepsilon/2 > 0$ there corresponds a natural $N_{1}\geq 0$ such that \begin{align*} n\geq N_{1} \Rightarrow d(x_{n},x) < \varepsilon/2 \end{align*}
Similarly, for every $\varepsilon/2$, there corresponds a natural number $N_{2}\geq 0$ such that \begin{align*} n\geq N_{2} \Rightarrow d(y_{n},x) < \varepsilon/2 \end{align*}
Gathering both results, we conclude that for every $\varepsilon/2 > 0$ there is a natural number $N = \max\{N_{1},N_{2}\}$ such that \begin{align*} n\geq N \Rightarrow |d(x_{n},y_{n}) - d(x,y)| & \leq |d(x_{n},y_{n}) - d(x_{n},y)| + |d(x_{n},y) - d(x,y)|\\\\ & \leq d(y_{n},y) + d(x_{n},x) < \varepsilon \end{align*} and we are done. Thus we conclude that any metric is continuous, since $d$ is arbitrary.
Is the wording of my proof correct? Should I fix any theoretical flaw?
Looks fine. You can also forgo the lemma, incorporating it in the proof of the result: for sufficiently large $n$ we have $d(x_n,x)<\frac{\epsilon}2$ and $d(y_n,y)<\frac{\epsilon}2$, so
$$d(x_n,y_n)\le d(x_n,x)+d(x,y)+d(y,y_n)<d(x,y)+\epsilon$$
and
$$d(x,y)\le d(x,x_n)+d(x_n,y_n)+d(y_n,y)<d(x_n,y_n)+\epsilon\;,$$
and therefore $|d(x,y)-d(x_n,y_n)|<\epsilon$.