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: 
I would say is as simple as that, since $\varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.
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.