What are the "technical troubles" with using a metric space rather than a topological space when defining an abstract manifold? (As in Spivak) One thing I think is interesting about Spivak's book A Comprehensive Introduction to Differential Geometry is that Spivak uses metric spaces instead of topological spaces when defining an abstract manifold.  For example, on p. 19: 

A manifold-with-boundary is a metric space $M$ with the following
  property: If $x \in M$, then there is some neighborhood $U$ of $x$ and
  some integer $n \geq 0$ such that $U$ is homeomorphic to either
  $\mathbb R^n$ or $\mathbb H^n$.

Munkres takes a similar approach in the final chapter of Analysis on Manifolds.
In the preface to the first edition of A Comprehensive Introduction to Differential Geometry, Spivak states:

An acquaintance with topological spaces is even better, since it
  allows one to avoid the technical troubles which are sometimes
  relegated to the Problems, but I tried hard to make everything work
  without it.

What are the "technical troubles" that Spivak is referring to here?  And why did Spivak have to try hard to make everything work out with this approach?  
It seems like building the theory using the metric space definition would be no more difficult than building the theory using the topological space definition.   I don't see what extra difficulties would arise.
 A: There are two major problems: If you remove the condition that $M$ is a metric space, then $M$ may not even be Hausdorff; the canonical example is $\mathbb{R}$ with a doubled origin. Furthermore, even if $M$ is Hausdorff, it may not be metrizable. 
In the topological setting, the usual requirement is that $M$ should be locally homeomorphic to some $\mathbb{R}^n$ (or $\mathbb{H}^n$, if we allow manifolds with boundary), and also that $M$ should be Hausdorff and second-countable. Hausdorff is generally a nice property have. Second-countability implies paracompactness and thus metrizability, and also implies that the manifold embeds in some $\mathbb{R}^N$ by a standard partition of unity argument.
I'm not familiar with how far Spivak gets into the subject, but dropping paracompactness does cause some technical problems. See, for example, Milnor and Stasheff's book on characteristic classes, which takes painful measures to avoid paracompactness for reasons that are really never explained in the text. In dealing with noncompact manifolds (compact manifolds are clearly paracompact), you can run into issues like the lack of an orientation class in the de Rham cohomology, failures of the existence of the normal bundle, etc. In particular, it can also prevent the existence of a Riemannian metric, which is probably what Spivak is referring to.
A: The main reason for not including a metric in the definition of differential manifold is that it is an irrelevant artificial structure which is immediately abandoned when one defines an atlas: obviously transition maps between charts are smooth but definitely not isometric!
In  the same vein the category of  smooth manifolds and that of metric spaces are not related by natural functors in either direction.
A more technical issue is that some manifolds are plain not metrizable, even in dimension one: Spivak himself describes an example, the long line (Appendix A, corollary 6).
It is true that such manifolds are not very common, but it is very unaesthetic to exclude them from the very definition, especially since you might start with metrizable manifolds and after operating on them with standard constructions end up with a non-metrizable one.
A: This arises mostly when you're constructing new manifolds. It's easy to define a quotient space; it's much harder to define a quotient metric space. (If the torus is $[0,1] \times [0,1]$ with appropriate edge identifications, how do you define the metric on the quotient?) 
This particular example is a bit disingenuous (see quotient metric space for how to get a metric on precisely this object) but the spirit is there.
It's often much easier to specify a topology than to specify a metric, and you will often want to build new manifolds. Of course, as noted in the comments in anomaly's answer, topological manifolds (meaning locally Euclidean, second countable, Hausdorff) are always metrizable; it's just that we don't want to have to do so much work to metrize them!
