What is the purpose of differentiating between norms and metrics? A metric $d(x,y)$ takes two points from some domain $X$ and returns a non-negative real number. It is the distance between two points.
A norm $n(x)$ takes only one point from $X$ and 
also returns a non-negative real number. It is the length of a vector when viewed as an arrow from the origin, or simply the distance between $x$ and $0$. 
My question is, why is there a need for distinguishing between norms and metrics when a norm is just a metric with one of its inputs being $0$? Is it correct to say that a norm is a type of metric, where one of the inputs is $0$?
 A: A norm induces a metric, using $d(x,y) = \|x-y\|$. But a norm is a function $X \to \mathbb{R}$ not $X \times X \to \mathbb{R}$, so in that sense it's simpler.
It only makes sense to talk about norms in a linear space, while any set can get a metric, no linear structure required. In a linear space we see that norms preserve "scale", when we multiply a vector by $2$ its norm is multiplied too, so the size of scalars is then coupled with the size of vectors.
There is a standard way to define a topology (and hence continuity etc.) from a metric space, from a norm we just go via the induced metric as well. So the metric is the more "fundamental structure", as it were. 
A: A metric is, in some sense, more general than a norm.  It might be helpful here to look at somewhat precise definitions:

Definition: Let $V$ be a vector space over $\mathbb{C}$ (or, perhaps more generally, over a valued field).  A norm on $V$ is a function $\|\cdot\| : V \to \mathbb{R}_{\ge 0}$ such that for any $x,y \in V$ and $a \in \mathbb{C}$, we have

*

*$\| x + y \| \le \| x \| + \| y \|$,

*$\| ax \| = |a| \| x \|$, and

*$\| x \| = 0$ if and only if $x = 0$.


Note that properties (1) and (2) depend on the vector space structure–in order to define a norm on a space, we have to be able to add and scale elements.  If you don't have a vector space structure, then you can't define a norm.  If the only spaces which you have ever seen are vector spaces, then this might seem like a silly condition, but it actually really does matter.
Next, a metric:

Definition: Let $X$ be a set.  A metric on $X$ is a function $d : X\times X \to \mathbb{R}_{\ge 0}$ such that for any $x,y,z\in X$, we have

*

*$d(x,y) \le d(x,z) + d(z,y)$,

*$d(x,y) = d(y,x)$, and

*$d(x,y) = 0$ if and only if $x=y$.


Note that we don't need a vector space structure on $X$ in order to define a metric.  Indeed, we don't need any structure at all–we can define a metric on any set at all.  For example, if $X$ is any set, then we could define $d(x,y) = 1$ whenever $x\ne y$ (and zero otherwise in order to satisfy (3)).
That being said, if you have a norm defined on a vector space, then you get a metric for free:  just define $d(x,y) := \| x - y \|$.  It is not that hard to verify that this does define a metric, though some argument is required.
Again, note that a norm is defined only on a vector space, while a metric can be more general.  For example, consider the usual ternary Cantor set.  We can measure distances between points in the Cantor set using the metric inherited from $\mathbb{R}$.  However, the Cantor set is not a vector space (at least, not as a subset of $\mathbb{R}$–there are vector spaces which are homeomorphic to the Cantor set as topological spaces), hence the metric does not come from a norm.  Indeed, we don't even have enough structure on this space in order to define a norm.
A: Of course one big difference is the context: We can talk about a metric on any set, while the notion of a norm only makes sense for a (real or complex) vector space.
So say $V $ is a vector space. If $||.||$ is a norm on $V$ and we define $d(x,y)=||x-y||$ then $d$ is a metric on $V$. And $d$ has two extra properties:
(i) $(V,d)$ is a topological vector space
(ii) $d$ is translation-invariant, meaning that $d(x+z,y+z)=d(x,y)$.
Let's temporarily call a metric on $V$ with those two properties a "vector metric".
So any norm  $||.||$  induces a vector metric $d$ via the definition $d(x,y)=||x-y||$, and as you note then $||x||=d(x,0)$.
But the notion of a vector metric on $V$ is still far from "the same as" a norm on $V$. Because if $d$ is a vector metric on $V$ and we define $||x||=d(x,0)$ it does not follow that $||.||$ is a norm.
