8 hours ago, by
philmcole I tried this: Take an open set $O \in T_1$ which means for every $x \in O$ there is an $\varepsilon$-neighborhood $B_\varepsilon(x)=\{y \in X \mid d_1(x,y) \lt \varepsilon\} \subseteq O$. Then by the equivalence of norms there are scalars $c,C$ s.t. $c d_1(x,y) \le d_2(x,y) \le C d_1(x,y)$.