If two spaces are homeomorphic and one is a metric space must the other be as well?

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.

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.