For the general question about how categorical concepts look when applied to metric spaces, one place to look is Lawvere's paper 'Taking categories seriously', section 6 onwards.
Aside from that, here are a few examples.
Functor categories became function spaces with the uniform or sup metric. That is, if $A$ and $B$ are metric spaces construed as enriched categories, then the functor category $B^A$ is the set of distance-decreasing maps $A \to B$ with the sup metric. (I use "decreasing" in the non-strict sense.)
The (cartesian) product $A \times B$ of two metric spaces --- that is, their product in the category of metric spaces --- has the '$\infty$-metric':
$$
d((a, b), (a', b')) = \max\{d(a, a'), d(b, b')\}.
$$
The same goes for infinite products --- remembering that $\infty$ is allowed as a distance. Once you know this, limits in general work in the obvious way.
The coproduct $A + B$ of two metric spaces $A$ and $B$ is their disjoint union, with $d(a, b) = d(b, a) = \infty$ for all $a \in A, b \in B$. Again, it's crucial here to allow $\infty$ as a distance. Otherwise, your category of metric spaces will lack lots of limits and colimits. The coequalizer of two maps $f, g: A \to B$ is $B$ quotiented out by the usual equivalence relation $\sim$ (as in the category of sets), and metrized by
$$
d([b], [b']) = \inf\{ d(y, y'): y \sim b, y' \sim b'\}
$$
where $[b]$ denotes the equivalence class of $b$. General colimits work similarly.
I mentioned the cartesian product, but there's another kind of product. Generally, if $\mathbf{V}$ is a monoidal category then any two $\mathbf{V}$-enriched categories, $A$ and $B$, have a tensor product $A \otimes B$. Its set of objects is the product of the sets of objects of $A$ and $B$. Its hom-objects are given by
$$
(A \otimes B)((a, b), (a', b')) = A(a, a') \otimes B(b, b').
$$
This gives us a tensor product of metric spaces. Given metric spaces $A$ and $B$, the point-set of $A \otimes B$ is the product of the point-sets of $A$ and $B$. The distance is given by
$$
d((a, b), (a', b')) = d(a, a') + d(b, b').
$$
In other words, it's the '$1$-metric', also known as the taxicab metric, Manhattan metric, etc.
So, Andrew, when you ask 'What is a monoidal metric space?', you have to say which product you want to be monoidal with respect to. That is, are you asking about (weak) monoids in $(\mathbf{Met}, \times)$ or in $(\mathbf{Met}, \otimes)$?
From the tone of your question, I would guess: both. So here goes.
The answer for cartesian product $\times$ doesn't seem so interesting. Assuming that your metric spaces satisfy the classical skeletality axiom ($d(a, b) = 0 \Rightarrow a = b$), a monoidal category for $\times$ is a metric space $A$ equipped with a monoid structure on its set of points such that
$$
d(a \cdot b, a' \cdot b') \leq \max\{d(a, a'), d(b, b')\}.
$$
I can't think of anything more to say about that.
The answer for tensor product $\otimes$ seems more interesting. A monoid in $(\mathbf{Met}, \otimes)$ is a metric space $A$ equipped with a monoid structure on its set of points such that for all $a$, the maps $a\cdot -$ and $- \cdot a$ are distance-decreasing. For example, if it's a group then this says that left or right translation is always an isometry. This often happens: consider the underlying additive group of a normed vector space, for instance.
