Metric is continuous, on the right track? 
Let $X$ be a metric space with metric $d$. Show that $d:X\times X\rightarrow \mathbb{R}$ is continuous.

The problem is taken from Munkres Topology second edition, Section 20.
I know that if $d$ is a metric on $X$ then $d:X\times X\rightarrow \mathbb{R}$. My thinking is that the topology that is on $X$ is the topology induced by the metric $d$, and that the topology on $X\times X$ is the product topology on that space where we take the basis to be
$$\mathcal{B}=\{ U \times V \mid \text{ $U,V$ both open in $X$}\}.$$
Am I on the right track to say that we define some new metric on $X\times X$ and show that this metric induces the same topology as the product topology and then work with the function $d$ as a function between metric spaces to show continuity?
The question doesn't mention anything about defining some new metric and I've tried to solve the problem by looking at $X\times X$ as having the product topology, but in picking some point $(x,y) \in X\times X$ and some neighborhood around $d(x,y)$ in $\mathbb{R}$, I haven't yet found the way to make a neighborhood around $(x,y)$ which maps into the neighborhood around $d(x,y)$.

To continue with the hint in the comments, prove that for any $x,y,z,w$
$$|d(x,y)-d(z,w)|\leqslant d(x,z)+d(y,w)$$
This shows that if $x\to x_n$ and $y\to y_n$ -- which is equivalent to $(x_n,y_n)\to (x,y)$ in $X\times X$ -- then $d(x,y)\to d(x_n,y_n)$, which proves that $d$ is continuous (since $X\times X$ is metric we can check continuity on sequences).
