There is no difference between a metrizable space and a metric space (proof included). Willard says, "Whenever $(X,\tau)$ is a topological space whose topology $\tau$ is the metric topology $\tau_{\rho}$ for some metric $\rho$ on $X$, we call $(X,\tau)$ a metrizable topological space."
I think giving a proof is the best way to illustrate how I think of these concepts and illustrate the exact points that I am not understanding.
Theorem: Metric space iff metrizable space.
(->) Let $(X,\rho)$ be a metric space. Consider the topology generated by this metric, $\tau_{\rho}$.  Then $(X,\tau_{\rho})$ is a topological space whose topology is the metric topology for some metric, so by definition, metrizable.
(<-) Let $(X,\tau)$ be a metrizable space.  There $\exists \rho$ a metric such that $\tau$ is the metric topology given by $\rho$.  And so $(X,\rho)$ is a metric space.
 A: Really formally: A topological space is a pair $(X, \mathcal{T})$ of a set $X$ and a subset $\mathcal{T}$ of the power set of $X$ satisfying the appropriate axioms for open sets. A metric space is a pair $(X, d)$ of a set $X$ and a map $d: X \times X \to \mathbb{R}^{\ge 0}$ satisfying the appropriate axioms for a distance function. It does not make sense to say a topological space "is" a metric space, or vice-versa, because there is an ontological status problem - how can $(X, \mathcal{T}) = (X, d)$ when $X$ and $\mathcal{T}$ are completely different objects? (See important caveat at bottom.)
If I give you a metric space $(X, d)$ then there is a canonical topological space $(X, \mathcal{T})$ with the same underlying set $X$, called the induced topological space, whose topology is generated by the $\epsilon$-balls.
If I give you a topological space $(X, \mathcal{T})$, there may or may not be some metric space $(X, d)$ whose induced topological space is $(X, \mathcal{T})$. If there is one, then we say that $(X, \mathcal{T})$ is metrizable. If $(X, \mathcal{T})$ is metrizable, then there is some metric $d$ such that $(X, d)$ is a metric space whose induced topological space is $(X, \mathcal{T}$, but note that for instance $(X, \frac{1}{2}d)$ also has this property, as does $(X, 7d)$, so there is absolutely nothing canonical about $d$, i.e. you can't recover $d$ from $\mathcal{T}$.
Caveat: Keeping $d$ and $\mathcal{T}$ around is horribly clunky and pretentious and nobody actually does it, and people say that a metric space "is a topological space" all the time, or vice-versa for metrizable topological spaces, e.g., I might say "the topological $\mathbb{R}^2$ is actually a metric space" when I mean "there exists a metric-space structure which induces the usual topology on $\mathbb{R}^2$." Being super-precise with ordered pairs is only useful when sorting out confusion like this one; it's not how to think about these things in practice.
A: A metric space is an ordered pair $\langle X,d\rangle$ such that $X$ is a set and $d$ is a metric on $X$. A metrizable space is an ordered pair $\langle X,\tau\rangle$ such that $X$ is a set, $\tau$ is a topology on $X$, and there exists a metric on $X$ that generates the topology $\tau$. These are clearly not the same thing. A metric space has a specified metric, and the topology, though definable from the metric, is unspecified; a metrizable space has a specific topology, and while it is possible to define metrics that generate that topology, none is specified.
