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 $\delta=\varepsilon/2$.
You have to prove:
$$\begin{eqnarray*}d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2 &\leq& d_X(x_1,x_2)^2+d_X(x_2,x_3)^2+d_Y(y_1,y_2)^2+d_Y(y_2,y_3)^2\\&+&2\sqrt{\left(d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2\right)\left(d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2\right)}\end{eqnarray*}$$
where the Cauchy-Schwarz inequality ensures that the last square root is greater or equal than:
$$ d_X(x_1,x_2)\,d_X(x_2,x_3)+d_Y(y_1,y_2)\,d_Y(y_2,y_3)$$
hence it is sufficient to show that:
$$d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq\left(d_X(x_1,x_2)+d_X(x_2,x_3)\right)^2+\left(d_Y(y_1,y_2)+d_Y(y_2,y_3)\right)^2$$
that just follows from the triangle inequality for $d_X$ and $d_Y$.
Best Answer
To prove that $P$ is continuous at $(x_1,y_1)$ you have to show that given $\epsilon >0$ there exists $\delta >0$ such that $d((x_1,y_1), (x_2,y_2)) <\delta$ implies $d_X(x_1,x_2) <\epsilon$. Take $\delta =\epsilon$ and use the inequality you have written.