How to show that $C=C[0,1]$ is a Banach space
Where do you get this subsequence $\{f_{n_k}\}_k$? Since a Cauchy sequence is convergent if and only if it has a convergent subsequence, you're essentially assuming the result is true here.
Here's a proof that $C[0,1]$ is complete (and thus a Banach space):
Suppose $\{f_n\}$ is Cauchy in $C[0,1]$. We must show that $f_n$ converges in the $C[0,1]$ norm to an $f$ in $C[0,1]$.
We first identify the "natural candidate" for $f$:
Since $\{f_n\}$ is Cauchy in $C[0,1]$, it follows that $\{f_n(x)\}$ is Cauchy in $\Bbb R$ for each $x\in[0,1]$. This observation, together with the fact that $\Bbb R$ is complete, gives us the well-defined function $f:[0,1]\rightarrow\Bbb R$ whose rule is $f(x)=\lim f_n(x)$.
Since the terms of $\{f_n\}$ get uniformly close to each other, we expect $f$ to be uniformly close to $f_m$ for large $m$:
Now let $\epsilon>0$ and choose $M$ so that $\|f_n-f_m\|_{C[0,1]}<\epsilon$ for $n, m\ge M$. Then for each $m>M$ and for any $x\in[0,1]$: $$ \tag{1} |f(x)-f_m(x)|=\lim_{n\rightarrow\infty}|f_n(x)-f_m(x)|\le \lim_{n\rightarrow\infty}\|f_m-f_n \|_{C[0,1]}\le\epsilon. $$
And, we finish up with some hand waving that should not seem arcane to someone studying Banach spaces:
From $(1)$, it follows that $f_n$ converges uniformly to $f$ on $[0,1]$. From this, it follows that $f\in C[0,1]$ (a uniform limit of continuous functions is continuous) and that $f_n$ converges to $f$ in $C[0,1]$.
Edit: A comment above leads me to remark:
$f$ is indeed continuous: Given $x\in[0,1]$ and $\epsilon>0$, choose $m$ so that $||f_n-f\,||_{C[0,1]}<\epsilon/3$ for $n\ge m$ and choose $\delta>0$ such that $|f_m(x)-f_m(y)|\le \epsilon/3$ for all $y$. Then if $|x-y|<\delta$: $$ |f(x)-f(y)| \le|f(x)-f_m(x)|+|f_m(x)-f_m(y)|+|f_m(y)-f(y)|<\epsilon. $$
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 $B(X)$.