shows that $d$ is continuous where $d$ is a metric defined on $X$ If $(X; d)$ is a metric space, then the metric $d$ on $X$ induces a product
metric $p$  on $X\times X$ by $p((x_1; y_1); (x_2; y_2)) := d(x_1; x_2) + d(y_1; y_2)$
Show that $d :(X \times X ) \rightarrow R$ is continuous. I know the definition of the epsilon delta definition but not sure how to show.

Let's assume $p$ is indeed a metric. Then we have a metric space $(X\times X,p)$ and a function $d:X\times X\to\mathbb R$, which we want to show is continuous with respect to $p$.
Let $(x,y)\in X\times X$ and $\varepsilon>0$. If $(x',y')\in X\times X$ then by definition,
$$p((x,y),(x',y'))=d(x,x')+d(y,y').$$
We want to make $|d(x,y)-d(x',y')|<\varepsilon$ by forcing the above to be small enough. If $p((x,y),(x',y'))<\delta$ then we know that both $d(x,x')<\delta$ and $d(y,y')<\delta$. This means that 
$$d(x,y)\le d(x,x')+d(x',y)<\delta+d(x',y)$$
and
$$d(x',y)\le d(x',x)+d(x,y)<\delta+d(x,y)$$
and so
$$|d(x,y)-d(x',y)|<\delta.$$
Similarly $|d(x',y')-d(x',y)|<\delta$. Thus,
$$|d(x,y)-d(x',y')|\le|d(x,y)-d(x',y)|+|d(x',y)-d(x',y')|<2\delta$$
and we can choose $\delta=\varepsilon/2$.
