Proof. Let \(\{f_n\}^{\infty }_{n=1}\) be Cauchy in \(C[0,1]\) and let \(\varepsilon >0\) be given. \(\exists N\in \mathbb {N}\ni m,n>N\\ \implies ||f_n-f_m||\leq \varepsilon \). So it therefore follows that \(\forall x\in [0,1]\) \[|f_n(x)-f_m(x)|\leq ||f_n-f_m||<\varepsilon \] Therefore, \(\{f_n(x)\}^{\infty }_{n=1}\) is Cauchy in \(\mathbb {R}\) and since \(\mathbb {R}\) is complete \(f_n(x)\rightarrow f(x)\in \mathbb {R}\).
Since this limit is unique for every \(x\in [0,1]\), it is a function on \([0,1]\). And since \([0,1]\) is compact and \(f_n:[0,1]\rightarrow \mathbb {R}\) is continuous \(\forall n\in \mathbb {N},\implies f_n\) is uniformly continuous on \([0,1]\). Thus \(f\in C[0,1]\).

So \(C[0,1]\) is complete \(\implies \) Banach space. □