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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?

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    ohh thank you for the help2012-07-22

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[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 $[-1,1]$.