If the question is asking you to prove it, then it's asking for you to apply the epsilon-delta definition of continuity. That is:
If f(x) is continuous at point p, then for any ε > 0 there exists δ > 0 such that if |p-x| < δ, then |f(p) - f(x)| < ε.
In order to prove this, consider 2 cases. First, if p ≠ 0:
For any ε > 0, we can choose δ = min(ε/2,|p|/2). Since both ε >0 and |p| > 0, then δ > 0, and for any x such that |p-x| < δ, we have
|p-x| < δ = ε/2 < ε
But then since δ < |p|/2, p and x are either both positive or both negative, so, |p-x| = ||p|-|x|| = |f(p)-f(x)|
Therefore,
|f(p)-f(x)| < ε
So far, our proof has been a long-winded way of proving that straight lines are continuous. For the other case, p = 0. Let ε > 0 and choose δ = ε/2. Then, for x such that
|p - x| < δ
|0 - x| < δ
|x| < δ
|f(x)-f(p)| = |f(x) - 0| = |f(x)| = |x| < δ = ε/2 < ε
Therefore, the function is also continuous at 0. Of course, you can clearly see this from the graph, but there's a lot of value in learning how to prove the obvious. If I lost you on any of the steps, please let me know.
Since your question mentions differentiability, note that the function is not differentiable at 0. The derivative will be:
f'(x) = {-1|x<0,1|x>0}
which is clearly not continuous (or defined) at 0. Therefore, the function is not differentiable at 0. It is differentiable everywhere else.
