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 $d(x,y)$.

Suppose $(X,d)$ is a metric space. Then the metric $d_*:(X\times X)\times(X\times X)\rightarrow \mathbb{R}$ defined by 
\begin{equation}
d_*\big((x,y),(x_o,y_o)\big)=\max\{d(x,x_o),d(y,y_o)\}
\end{equation}
is a metric on $X\times X$ induce by the metric $d$ on $X$.
Now, let $(x_0,y_0)\in X\times X$ and $\epsilon>0$. If $(x,y)\in X\times X$ and $d_*\big((x,y),(x_0,y_0)\big)<\frac{\epsilon}{2}$. Then 
\begin{align*}
|d(x,y)-d(x_0,y_0)| &=|d(x,y)-d(x_0,y)+d(x_0,y)-d(x_0,y_0)|\\
     &\leq |d(x,y)-d(x_0,y)|+|d(x_0,y)-d(x_0,y_0)|\quad (\text{By Triangular Inequality})\\
     &\leq d(x,x_0)+d(y,y_0)\quad \quad (\text{By  Reverse Triangule Inequality})\\
     &\leq \max\{d(x,x_o),d(y,y_o)\}+\max\{d(x,x_o),d(y,y_o)\}\\
     &= d_*\big((x,y),(x_0,y_0)\big)+d_*\big((x,y),(x_0,y_0)\big)\\
     &<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}
Hence, $d$ is continuous.
