We are given two metric spaces $(\mathfrak X_1,d_1)$ and $(\mathfrak X_2,d_2)$ which we assume separable and such that $\mathfrak X:= \mathfrak X_1 \cap \mathfrak X_2 \neq \emptyset$.
Define $d(x,y):= d_1(x,y) + d_2(x,y)$ for $x,y\in\mathfrak X$.
Is $(\mathfrak X,d)$ separable?
My attempt: let $(u_n)_n$ be dense in $\mathfrak X_1$, and $(v_m)_m$ be dense in $\mathfrak X_2$. Pick rational radii $0<q\in\mathbb Q$. Then pick $y_{n,q}\in B(u_n,q;\mathfrak X_1)\cap \mathfrak X_2$ and $z_{m,q}\in B(v_m,q;\mathfrak X_2)\cap \mathfrak X_1$ , whenever these intersections are not empty. That would be the candidates for my dense set, but I cannot conclude from here.
A metric space is separable iff it is second-countable. So the assumption is that we have two metrics $d_1,d_2$ on $X$ that do not necessarily have any relation, except that the topologies $T_1,T_2$ they induce on $X$ are both second-countable. The topology $T$ induced by $d_1+d_2$ is exactly the topology generated by $T_1,T_2$, i.e, the smallest topology that contains them. A basis for $T$ is $\{U\cap V\mid U\in T_1,V\in T_2\}$. You can check that this is homeomorphic to the diagonal $D=\{(x,x)\mid x\in X\}$ in $(X,T_1)\times(X,T_2)$. Since $(X,T_1)\times(X,T_2)$ is second-countable, so is $D$, and thus $(X,T)$. Therefore $(X,T)$ is separable.