If X can be topologically embedded in a metric space Y how would one define the metric on X?

Let $X$ be a topological space and $i: Xto Y$ a topological embedding into a metric space $(Y, d)$.

The most important thing to note is that $S$ is open in $X$ if and only if $f(S)$ is open in $Y$. Equivalently, $S$ is a neighborhood around $xin X$ if and only if $f(S)$ is a neighborhood around $f(x)in Y$. Therefore, intuitively speaking, having an embedding means that we can effectively treat $X$ as a subspace of $Y$.

So to define a metric $delta$ on $X$, we can transport our points to $Y$ and measure the distance there: $$ delta(x, x') := d(f(x), f(x')). $$ Proving that this indeed is a metric is essentially just transporting the metric axioms on $Y$ through the injection $f$.

To show that the topology induced by $(X, delta)$ is the same as the original one, we can just verify that the notion of „neighborhood“ is the same (as it is purely definable by the notion of open sets, hence only depends on the topology): note that the following are equivalent for $xin X, Ssubseteq X$:

1. $S$ is a neighborhood around $x$
2. $f(S)$ is a neighborhood around $f(x)$
3. $f(S)$ contains an $epsilon$-ball around $f(x)$ with regards to $d$
4. $S$ contains an $epsilon$-ball around $x$ with regards to $delta$

Edit: Your title said „topological embedding“, and your post just spoke of „homeomorphic“. Therefore I chose to show the more general principle.