Why do we have the notions of both 'norm' and 'metric'? In the case of vectors in euclidean space, for instance, we can express one in terms of the other--i.e. length is distance from zero, distance is the length of the vector difference.  Does this break down somewhere?
 A: Compare Normed space with Metric space.
A metric space is weaker, all normed vector spaces are metric spaces.
However, there are metric spaces that are not normed.
The thing that fails, is the property $|\alpha x| = |\alpha| |x|$
which is only required in a vector space, but not in normed spaces.
As an example, define $d(x,y) = |x-y|/(1+|x-y|)$ as metric on the reals.
I believe this should satisfy the triangle inequality, and the other properties are clear. It is evident that $d(\alpha x,0) \neq \alpha d(x,0)$ so this is no norm.
A: This is basically what Mark Schwarzmann was able to say in one sentence in his comment:

In a vector space (over R , say), you can talk about "positive homogeneous translation-
  invariant metrics". "Norm" just sounds a bit nicer.

Let me elaborate a little bit.
It is correct that we could define linear normed space as a vector space with a metric on it, which fulfills some conditions. (Basically, by replacing $\lVert x \rVert$ by $d(x,0)$ in definition of norm; but probably various other definitions would be possible.) This would be an equivalent definition. However, it would be cumbersome to write $d(x,0)$ always instead of $\lVert x \rVert$, wouldn't it. Also, if we give a name to some concept, it is the first step to thinking about it.
Generally, if one concept can be reduced to another one (defined in terms of another concept), it doesn't mean it is redundant. It might not only bring brevity, but also more insight and possibilities to generalize it.
Just one simple example, which comes to my mind, since I used it recently. A topological space is compact $\Leftrightarrow$ every open cover has a finite subcover $\Leftrightarrow$ every system of closed sets having finite intersection property has a non-empty intersection.
Although the transformation between these two equivalent conditions is easy (basically de Morgan's law) and every proof using one of them could be rewritten to the proof which uses second one almost mechanically; in some situations one of them is more suitable and in some situations the other one works better. If we chose to use always only one of them, it would sometimes lead to less clear proofs.

I hope this is more-or-less what your question was about. Sorry, if it was too much digression from what you wanted.
