Metric Topology Same as Subspace Topology? 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.
 A: If $d_Y$ is the metric of $X$ restricted to $Y$ (so essentially $d\restriction_{Y \times Y}$), the main observation is that for the balls in this metric we have, almost by definitions:
$$\forall y \in Y: \forall r>0: B_{d_Y}(y,r) = B_d(y,r) \cap Y$$
so that the $d_Y$-balls in $Y$ are relatively open in $Y$, which already implies that the metric topology on $Y$ induced by $d_Y$ is a subset of the subspace topology, and on the other hand if $O$ is open in $X$, and $y \in O \cap Y$, some $d$-open ball around $y$ sits inside $O$ and then the intersected ball (thus a $d_Y$ ball by the above) sits inside $O \cap Y$, making all members of $O \cap Y$ interior points in the $d_Y$-topology, and so the subspace topology is a subset of the topology induced by $d_Y$. Hence the equality of topologies.
A: Let $U$ be open within $Y$.
Thus there exists $V$ open within $X$ with $U = V \cap Y$.
For all $x$ in $V$, exists ball $B(x)$ with $B(x) \subseteq V$.
$V = \bigcup \{ B(x) : x \in V \}$.
$$U = V \cap Y = \bigcup \{ B(x) : x \in V \} \cap Y =   
\bigcup \{ B(x) \cap Y : x \in V \} = \bigcup \{ B(x) \cap Y : x \in X \}$$
(The last equality requires some thought to prove.)
As every open set of $Y$ is the union of a collection of balls of $Y$,
$Y$ is metrized by the metric of $X$.
