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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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As Pete points out in his answer the idea can be made into an argument (although I myself wouldn't apply Stone-Weierstrass, that's quite an overkill for this question). However, you should pick another function, e.g. $x \mapsto |x-1/2|$ (which you can approximate uniformly by explicit polynomials -- how?) Note that the function $|x| = x$ on $[0,1]$ is $C^1$. –  t.b. Jul 22 '12 at 13:36
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Note that, uniform convergence preserves continuity, but it does not preserve differentiability. –  Mhenni Benghorbal Jul 22 '12 at 13:39
    
so my argument is incorrect in $[0,1]$ :( but I dont know how to prove the result Mr. Pete has written in his answer –  El Angel Exterminador Jul 22 '12 at 13:45
    
@Patience: Switching from $|x|$ to (say) $|x-\frac{1}{2}|$ seems like a very minor adjustment. Is there something else in my answer that you don't understand? –  Pete L. Clark Jul 22 '12 at 13:58
    
how to show $C[0,1]$ is dense in $C^1[0,1]$? –  El Angel Exterminador Jul 22 '12 at 14:08

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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 $[-1,1]$.

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