If two sequences converge in a metric space, the sequence of the distances converges to the distance of limits of the sequences. Let $(p_n)$ and $(q_n)$ be sequences in the metric space $(X, d)$ and assume that $p_n \rightarrow p \in X$ and $q_n \rightarrow q \in X$. Prove that $d(p_n, q_n)$ converges to $d(p, q)$.
Ok, so using the triangle inequality (and assuming the sequences are Cauchy - but can I do that?), I can prove that the distance does converge, but how do I say it converges to $d(p,q)$ exactly?

The key is to use the reverse triangle inequality. Since $d$ is a distance, you have $d(x,y)\leq d(x,z)+d(z,y)$ for any $x,y,z\in X$. This you can write as 
$$
d(x,y)-d(z,y)\leq d(x,z).
$$
As the roles of $x$ and $z$ can be reversed, you get the reverse triangle inequality
$$
|d(x,y)-d(z,y)|\leq d(x,z).
$$
Then 
$$
|d(p_n,q)-d(p,q)|\leq d(p_n,p),\ \ \ 
|d(p_n,q_n)-d(p_n,q)|\leq d(q_n,q).
$$
Using the above and the regular triangle inequality,
$$
|d(p_n,q_n)-d(p,q)|\leq |d(p_n,q_n)-d(p_n,q)|+|d(p_n,q)-d(p,q)|\leq d(q_n,q)+d(p_n,p).
$$
So $\lim_{n\to\infty}d(p_n,q_n)=d(p,q)$.
