I've thought about this question for a while, from Basener's Topology and its Applications.
Let $X$ be a metric space with a metric $d$ and $Y \subseteq X$. Prove that the subspace topology on $Y$ inherited from $X$ is the same as the metric topology from the metric $d$ on $Y$.
Here's an attempt that seems fuzzy: Choose $U \subset Y$ open. Then $U$ is open in $X$ under the subspace topology. But one can construct a ball $B(x)$ with some radius $r$, where $x \in U$ using the metric $d$, and so $U$ is open in $X$ in the $d$-metric topology.
My thought is that two topologies are the same if the open sets of one are also open in the other. I am unsure, however, if my approach is correct.