Question: Is every metric space a subspace of some connected metric space? If the space itself is connected then we're done, but if not then I think we can extend our metric space to make it connected. I'm not sure whether this will work or not, but intuitively I think the answer is yes.

Answer: 
In fact every metric space admits a canonical isometric embedding into a contractible metric space, as follows. If $X$ is a metric space, let $\hat{X}$ denote the space of weak contractions $X \to \mathbb{R}\_{\ge 0}$, where a *weak contraction* $X \to Y$ between two metric spaces is a map $f : X \to Y$ such that $d\_Y(f(x\_0), f(x\_1)) \le d\_X(x\_0, x\_1)$. This space is a metric space when equipped with the sup norm

$$d\_{\hat{X}}(f, g) = \sup\_{x \in X} |f(x) - g(x)|.$$

There is a canonical embedding

$$X \ni x \mapsto (y \mapsto d(x, y)) \in \hat{X}$$

which one can straightforwardly verify is an isometry. This is a slight variant of the [Yoneda embedding](http://ncatlab.org/nlab/show/Yoneda+embedding) for metric spaces regarded as [enriched categories](http://ncatlab.org/nlab/show/metric+space#lawvere_metric_spaces).
Here is another way to embed any metric space $X$ in a connected metric space. For each pair of distinct points $x,y\in X$, adjoin to $X$ a path from $x$ to $y$, and extend the metric to points on the path by making the path have length $d(x,y)$ (so distances are defined by taking the least distance which can be achieved along the paths or through $X$; by making the paths have length $d(x,y)$, distances between points of $X$ will be unchanged). The result will be a metric space in which $X$ embeds isometrically which is connected (in fact, path-connected).