How to show that a subspace of a metrizable space is also metrizable?
Let $X$ be a topological space, and let $Y$ be a (non-empty) subset of $X$; let $d$ be a metric on $X$ that induces the topology of $X$. Then the restriction $d | Y \times Y$ of $d$ to the subset $Y \times Y$ of $X \times X$, which is defined by $$ \big( d | Y \times Y \big) ( p, q) := d(p, q) \ \mbox{ for all } p, q \in Y, $$ is also a metric on $Y$.
Now let $\mathscr{T}$ be the subspace topology that $Y$ inherits as a subspace of $X$, and let $\mathscr{T}^\prime$ be the topology induced by the metric $d | Y \times Y$ on $Y$.
How to show that these two topologies on $Y$ are the same?
My Attempt:
Let $V$ be a $\mathscr{T}$-open set of $Y$, and let $y \in Y$. Then there exists an open set $U$ of $X$ such that $V = Y \cap U$, and our $y \in U$ of course, which implies the existence a real number $\epsilon_y > 0$ such that $B_d \left( y, \epsilon_y \right) \subset U$, that is, for any point $x \in X$ such that $d(x, y) < \epsilon_y$, we also have $x \in U$; therefore for any point $x \in Y \subset X$ in particular such that $\big( d | Y \times Y \big) (x, y) < \epsilon_y$, we have $x \in U$ and hence $x \in Y \cap U = V$, which implies that $$ B_{d | Y \times Y} \left( y, \epsilon_y \right) \subset V, $$ and so $$ V = \bigcup_{y \in V} B_{d | Y \times Y} \left( y, \epsilon_y \right), $$ from which it follows that $V$ is also a $\mathscr{T}^\prime$-open set and hence that $$ \mathscr{T} \subset \mathscr{T}^\prime. $$
Am I right?
Conversely, let $V^\prime$ be a $\mathscr{T}^\prime$-open set of $Y$, and let $y^\prime \in V^\prime$. Then there exists a real number $\epsilon_{y^\prime} > 0$ such that $$ B_{d | Y \times Y } \left( y^\prime, \epsilon_{y^\prime} \right) \subset V^\prime, $$ that is, for any point $y \in Y$ such that $\big( d | Y \times Y\big) \left( y, y^\prime \right) < \epsilon_{y^\prime}$, we also have $y \in V^\prime$. Let us put $$ U^\prime := \bigcup_{ y^\prime \in V^\prime} B_d \left( y^\prime, \epsilon_{y^\prime} \right). $$ This set $U^\prime$ is an open set of $X$ of course and $V^\prime = Y \cap U^\prime$, thus showing that $V^\prime$ is $\mathscr{T}$-open, which implies that $$ \mathscr{T}^\prime \subset \mathscr{T}. $$
Am I right?
Is the above proof correct, clear, and complete enough in each and every detail? Or, are there any gaps therein?