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: 
If $(X, d\_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.

The metric is indeed a transportation of the given one on $X$: $\phi: X \to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $\psi: Y \to X$ so that $\psi \circ \phi = 1\_X$ and $\phi \circ \psi = 1\_Y$), and we can define $$d\_Y: Y \times Y \to \mathbb{R} \text{ by: } d\_Y(y\_1,y\_2) = d\_X(\psi(y\_1), \psi(y\_2))$$

One easily checks the axioms for a metric for $d\_Y$ and by bijectiveness and the definitions we get that $\phi[B\_{d\_X}(x, r)] = B\_{d\_Y}(\phi(x), r)$ for all $x \in X, r>0$ etc. so that balls in the $d\_X$ metric get mapped to balls in the $d\_Y$-metric showing (with a little thought) that indeed open balls under $d\_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.
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.