Question: So I've been asked to prove non-differentiability using this particular method, and I'm a bit lost. I'm supposed to prove that $f(x,y) = |x-y|$ is not differentiable along the $x=y$ line. To do it, I'm supposed to use this definition of differentiability:

A function $f$ is called differentiable at a point $a$ in it's domain if there exists a vector $c\in\mathbb{R^{n}}$ such that:

$$\lim\limits\_{\bf h \to \bf 0} \frac{f(\boldsymbol a + \boldsymbol h) - f(\boldsymbol a) - \boldsymbol c \cdot \boldsymbol h}{|h|} = 0$$

Okay, so obviously $c$ would be the gradient of $f$ at that point, and obviously it won't exist because there is an apex there, but I can't for the life of me prove that it doesn't exist. Any help would be appreciated.

Answer: 
A subset $A$ of a set $X$ is closed with respect to a metric $d$ on $X$ precisely if every point in the closure of $A$ is in $A$. A point $a\in X$ is in the closure of $A$ if for every positive number $\varepsilon$, no matter how small, the ball of radius $\varepsilon$ centered at $a$ intersects $A$.
An alternative but very important way to define a closed subset of a metric space X is as the complement of an open subset of X. The definition of an open subset is defined here:

<http://en.wikipedia.org/wiki/Open_set#Metric_spaces>

You can then define a closed subset A of X as any set where it's complement X\A is open in X.

This definition of closed subset of X is equivalent to Michael's definition (and as an exercise you should try to prove that the two definitions are equivalent).