Correct answer: (A) Convergence in $(X,d_2)$ implies convergence in $(X,d_1)$.
Explanation
The topology $\mathcal T_2$ is finer, so every $\mathcal T_1$ neighbourhood is also a $\mathcal T_2$ neighbourhood. A sequence eventually inside every $\mathcal T_2$ neighbourhood is therefore eventually inside every $\mathcal T_1$ neighbourhood. Thus A. The continuous identity goes from the finer topology to the coarser one, and equality of open sets does not require equality of metric balls.
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