I need to show that the product of two metrizable topological space $(X, T_1)$ and $(Y, T_2)$ is metrizable and the metric is given $d=d_1+d_2$ where $d_1$ is the metric that induces the topology $T_1$ and $d_2$ induces $T_2$.
I guess showing that product of $(X, d_1)$ and $(Y, d_2) = ((X,Y), d)$. I'm not sure where to start, but I guess we need to show any open ball in the product metric $B((x,y),r)$ contains the product of open balls from $B(x,r)$ and $B(y,r)$ and vice versa.
First note that the projections $\pi_X: X \times Y \to X$ and $\pi_Y: X \times Y \to Y$ are both continuous seen as maps from $(X \times Y,d)$ because e.g.$$d_1(\pi_X((x,y),\pi_X(x',y')) = d_1(x,x') \le d((x,y),(x',y'))$$ so the projections are contractions (so even uniformly continuous, $\delta=\varepsilon$ works globally).
Because the product topology $\mathcal{T}_p$ on $X \times Y$ is defined as the smallest topology on $X \times Y$ that makes both $\pi_X$ and $\pi_Y$ continuous, and the metric topology from $d$ is one such topology, we have that $\mathcal{T}_p \subseteq \mathcal{T}_d$, showing one inclusion.
To show the reverse, take any open set $O$ in $\mathcal{T}_d$ and $(p,q)$ in $O$. We have to show $(p,q)$ is an interior point for $O$ in the product topology. As the point is arbitrary, it follows that $O$ is then product-open and we're done showing the equality of these topologies.
First note that is an open ball of radius $r>0$ around $(p,q)$ such that $B_d((p,q),r) \subseteq O$ because $O$ is $d$-open. Then note that $U=B(p,\frac{r}{2})$ is open in $X$ and $V=B(q, \frac{r}{2})$ is open in $Y$ (being metric balls in the metric topology) and we claim that $(p,q) \in U \times V \subseteq B_d((p,q), r)$, which is quite obvious from the definitions. But $U \times V$ is product open and so indeed $(p,q)$ is an interior point of $O$ for the product topology.
