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 $d_Y$.
There are two things to be done:
a) Define a metric on $X \times Y$ (based on $d_X$ and $d_Y$). The product of the metrics won't work (as then the distance is already $0$ for points like $(x,y)$ and $(x,y')$, to name an easier reason), but the sum or max both work and are convenient choices. So e.g. (my favourite choice for 2 spaces) define
$$d((x_1, y_1), (x_2, y_2)) = \max(d_X(x_1, x_2), d_Y(y_1, y_2))$$
and check that this indeed gives a metric on $X \times Y$.
b) Secondly, and quite importantly, one has to check that the topology of $(X \times Y, d)$ is actually the same as the product topology on $X \times Y$. That will show that the product of $X \times Y$, which by definition/convention will have the product topology, can also be induced by a metric, i.e. is metrisable.
E.g. check that $$B_d((x,y), r) = B_{d_X}(x,r) \times B_{d_Y}(y, r)$$
which shows that open balls under $d$ are indeed open in the product topology of $(X,d_X)$ and $(Y,d_Y)$, so $\mathcal{T}_d \subseteq \mathcal{T}_{\text{prod}}$. The reverse also needs showing: suppose $O$ is product open, and let $(x,y) \in O$. Then there is an open set $U$ in $\mathcal{T}_X$ (topology induced by $d_X$) and an open set $V$ in $\mathcal{T}_Y$, such that
$(x,y) \in U \times V \subseteq O$. This means that there is some $r_1>0 $ such that $x \in B_{d_X}(x,r_1) \subseteq U$ and some $r_2>0$ such that $y \in B_{d_Y}(y,r_2) \subseteq V$. But then, setting $R =\min(r_1, r_2)>0$:
$$(x,y) \in B_d((x,y), R) = B_{d_X}(x,R) \times B_{d_Y}(y,R) \subseteq B_{d_X}(x,r_1) \times B_{d_Y}(y,r_2) \subseteq U \times V \subseteq O$$
showing $O \in \mathcal{T}_d$ as required. So the product topology coincides with the topology generated by $d$.
As a final remark: one can show that this extends to countable products of metric spaces as well. This is somewhat more work, and then a sum metric is the most suitable:
$$d((x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}}) = \sum_{n \in \mathbb{N}} \frac{1}{2^n}\min(d_{X_n}(x_n, y_n), 1)$$
will then do as a metric on $\prod_n X_n$ that induces the product topology. See this answer for more details.
Also s
Best Answer
I'm afraid @Mindlack is right. Note that $$|x_1-x_2|^2+|x_2-x_3|^2-|x_1-x_3|^2\\=x_1^2+x_2^2-2x_1x_2+x_2^2+x_3^2-2x_2x_3-x_1^2-x_3^2+2x_1x_3\\=2(x_2-x_1)(x_2-x_3)$$can be positive, negative or zero depending on the values of $x_1,\,x_2,\,x_3$. A similar result holds for the $y_i$. Thus $$|x_1-x_2|^2+|x_2-x_3|^2-|x_1-x_3|^2+x_i\mapsto y_i$$needn't be non-negative, as the triangle inequality would require. In fact, we can also think of this in terms of the cosine rule in planar Euclidean geometry: if $x_1,\,x_2,\,x_3$ are the vertices of a triangle with internal angle $\theta$ at $x_2$,$$|x_1-x_2|^2+|x_2-x_3|^2-|x_1-x_3|^2=2|x_1-x_2||x_2-x_3|\cos\theta$$has the same sign as $\cos\theta$.