Question: Suppose $X$, and $Y$ are topological spaces, and that they are homeomorphic. If $X$ is a metric space must $Y$ be as well?

Here is my thought process.

If $X$ is a metric space then there exists some distance function on $X$ given by $d:X\times X\rightarrow\Bbb R$. Let $\varphi:X\rightarrow Y$ be a homeomorphism between $X$ and $Y$. 

Then for any distinct points $y\_1,y\_2\in Y$ we can determine a "distance" between them by $d(\varphi^{-1}(y\_1),\varphi^{-1}(y\_2))\in\Bbb R$

To be a metric space however, I need some way of combining this idea of passing the points back to $X$ from $Y$ in a nice way, and then applying the distance function on $X$. Would it be as simple as defining $d'(y\_1,y\_2) = d(\varphi^{-1}(y\_1),\varphi^{-1}(y\_2))$?

If not is there a way to do that? Is this enough to say that $Y$ is a metric space? Or am I just out of my mind?

Answer: 
You construction can be done regardless of $\varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(\varphi^{-1}(-),\varphi^{-1}(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y \in Y$ and $\varepsilon > 0$ . Now, 

$$
B'\_\varepsilon(y) := \{x:d'(x,y) < \varepsilon\} = \{x : d(\varphi^{-1}(x), \varphi^{-1}(y)) < \varepsilon\} = \varphi(B\_\varepsilon(\varphi^{-1}(y))).
$$

Since $B\_\varepsilon(\varphi^{-1}(y))$ is a ball in $X$ and $\varphi$ is homeo, $B'\_\varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U \subseteq Y$ is open for the original topology,

$$
U = \varphi(\varphi^{-1}(U)) = \varphi(\bigcup\_{i \in I}B\_{r\_i}(x\_i)) = \bigcup\_{i \in I}\varphi(B\_{r\_i}(x\_i)) = \bigcup\_{i \in I}B'\_{r\_i}(\phi(x\_i))
$$

and so $U$ is open for the metric topology. Here we use that $\varphi^{-1}(U)$ is an open set of $X$ and so it is a union of open balls.
Given any sets A and B and a bijection $\varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b\_1+b\_2$ as being $\varphi(\varphi^{-1}(b\_1)+\varphi^{-1}(b\_2))$, then $\varphi$ will be a group isomorphism with respect to addition in A and addition in B. 

Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ *along with* its distance function is a metric space, and $Y$ along with the metric induced by $\varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.