TITLE: Is this metric space complete?
QUESTION [5 upvotes]: No, it is not complete metric space: by Stone-Weierstrass thm we know that $|x|$ can be uniformly approximated by sequence of polynomials which are clearly $\mathcal{C}^1[0,1]$, but $|x|$ is not $\mathcal{C}^1$.
Is my argument correct?

REPLY [3 votes]: [Added: By $C[0,1]$ I mean the set of continuous functions $f: [0,1] \rightarrow \mathbb{R}$ endowed with the metric $d(f,g) = \max_{x \in [0,1]} |f(x)-g(x)|$.  This is a complete metric space by the Cauchy Criterion for Uniform Convergence.]
Yes.  To recap it: you have a complete metric space, $\mathcal{C}[0,1]$, and a subspace, $\mathcal{C}^1[0,1]$, which is not closed (rather, it is proper and dense).  Therefore $\mathcal{C}^1[0,1]$ cannot be complete.  
Added: As t.b. points out, the absolute value function is $C^1$ on the interval $[0,1]$, so you should pick something else (e.g. what t.b. says).  I also agree that Weierstrass Approximation is much more than you need here.  For instance, in Example 7 of these notes I show -- in an intentionally clunky, hands-on fashion -- that the absolute value function is a uniform limit of $C^1$-functions on $[-1,1]$.