TITLE: How to show that $C=C[0,1]$ is a Banach space
QUESTION [20 upvotes]: Let $C=C[0,1]$ be the space of all continuous functions on $[0,1]$. Define $\|f \|=\max \ |f(x)|$. I want to show that $C$ is a Banach space.  

Below is my attempt and I was wondering if it's ok.  
I know I have to show that $C$ is a complete normed space.
Clearly, $\|f\| \geqslant 0$ and $\|f\|=0 \Leftrightarrow f=0$. $\|cf \|=\max~|cf(x)|=|c|\max |f(x)|=|c| \cdot \|f\|$.
$\|f+g\|=\max~|f(x)+g(x)|\leq \max~|f(x)|+\max~|g(x)|=\|f\|+ \|g\|$.
So $C$ is a normed space. 
Next, I show that every Cauchy sequence in $C$ is convergent.
Let $\{f_n\}$ be a Cauchy sequence in $C$.
Let $\varepsilon \gt 0.$ Then $\exists$ an $N_1$ such that $$ \max~|f_n(x)-f_m(x)| \lt \frac{\varepsilon}{2}$$
for $n, m \gt N_1$ and $x\in[0,1]$.
But there is a subsequence $f_{k_n} $, which converges to $f$. So $\exists$ an $N_2$ such that
$$ \max~\left|f_{k_n} - f\right|\lt \frac{\varepsilon}{2}$$
for each $n\gt N_2$.
Now Let $N = \max\{N_1, N_2\}$, if $n \gt N$ then $k_n \geqslant n\gt N$. So we have
$$ \max~\left|f_n(x) - f(x)\right| \leqslant \max~\left|f_n - f_{k_n}\right| + \max~\left| f_{k_n} - f\right| \lt\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$ 
Thus, $\|f_n-f\| \to 0$ as $n\to \infty$. $\quad \square$
Thanks.

REPLY [2 votes]: The usual way to prove that $X$ compact Hausdorff $\implies C(X,\mathbb K)$ is a Banach space over $\mathbb K$ goes in two steps:


Show that $B(X) := \ell^\infty(X,\mathbb K) = \{f \in \mathbb K^X \mid f$ is bounded $\}$ is a Banach space w.r.t. the sup-norm.
Show that the uniform limit of a sequence of functions that are continuous at a point is continuous at that point as well. From this it follows that $C(X)$ is closed in $B(X)$.