So we want to prove that $C[a,b]$ is a complete metric space under the metric $d(f,g) = ||f-g||_\infty = \sup_{x\in [a,b]}|f(x)-g(x)|$. To do this, we need to prove that for any Cauchy sequence $\{f_n\}$, where all the $f_n \in C[a,b]$, they converge to some $f\in C[a,b]$; putting this in math notation, we want to show that if for all $\epsilon>0$, there is $N \in \mathbb N$ such that $m,n\geq N \implies d(f_m,f_n)<\epsilon$, then there is some $f\in C[a,b]$ such that for all $\epsilon>0$, there is $N \in \mathbb N$ such that $n\geq N \implies d(f_n,f)<\epsilon$.
We can show the existence of such an $f$ by directly constructing such an $f$: take $f$ to be the pointwise limit of the $f_n$.
Step 1: We have to show that we can actually take the pointwise limit of the $f_n$. Ok; take any $x_0\in [a,b]$. Then, for any $n,m$, $|f_n(x_0)-f_m(x_0)| \leq d(f_n,f_m)$ which is $<\epsilon$ for $n,m$ sufficiently large. That means that $\{f_n(x_0)\}$ is a Cauchy sequence (of real numbers), and hence converges to some value (see: http://mathonline.wikidot.com/the-cauchy-convergence-criterion), which we'll call $f(x_0)$.
Step 2: We need to verify that $f$ is actually in $C[a,b]$; this is true because the $f_n$ converge uniformly to $f$ (a fact we will prove in Step 3) and the $f_n$ are continuous. Details are in step (iii) of Is the space $C[0,1]$ complete?.
Step 3: Now that we have this candidate function $f$, let us prove that for all $\epsilon>0$, there is $N \in \mathbb N$ such that $n\geq N \implies d(f_n,f)<\epsilon$. So we've fixed some $\epsilon>0$ and found the corresponding $N\in \mathbb N$ s.t. $m,n\geq N \implies d(f_m,f_n)<\frac \epsilon3$; then
$$d(f_n,f) \leq d(f_n,f_N) + d(f_N, f) < \frac \epsilon 3 + d(f_N,f) \qquad (0)$$
We now have to develop a bound for $d(f_N,f)$. We do this pointwise. For any $x_0\in [a,b]$, we have for any $m\geq N$:
$$|f_N(x_0) - f(x_0)| \leq |f_N(x_0)-f_m(x_0)| + |f_m(x_0) - f(x_0)| \qquad (1)$$
Well we know that for any $m\geq N$, $|f_N(x_0)-f_m(x_0)|\leq d(f_N, f_m) < \frac \epsilon 3$ (so the first part of this bound comes from the uniform bound). But we also know that since $\lim_{m\to\infty} f_m(x_0) = f(x_0)$, we can make $|f_m(x_0) - f(x_0)|$ as small as we want by choosing sufficiently large $m$. We can in fact choose $m$ as big as we want because recall the above inequality $(1)$ holds "for any $m\geq N$", and so we can definitely choose $m$ large enough to force $|f_m(x_0) - f(x_0)| <\frac \epsilon 3$. Thus, we can rewrite $(1)$ to be
$$|f_N(x_0) - f(x_0)| \leq |f_N(x_0)-f_m(x_0)| + |f_m(x_0) - f(x_0)| < \frac \epsilon 3 + \frac \epsilon 3 \qquad (1')$$
The magic of this part of the proof is that we split $|f_N(x_0) - f(x_0)|$ into two pieces involving the parameter $m$, where $m$ was chosen cleverly to get a bound on those two pieces NOT involving $m$!
To summarize, in this step, we proved that for all $x_0\in [a,b]$, $|f_N(x_0) - f(x_0)| < \frac {2\epsilon}{3}$. That means that we can take the supremum over ALL $x_0 \in [a,b]$ and keep a weak version of the bound:
$$d(f_N, f) = \sup_{x_0 \in [a,b]} |f_N(x_0) - f(x_0)| \leq \frac {2\epsilon}{3}$$
Putting this with $(0)$, we get that
$$d(f_n,f) \leq d(f_n,f_N) + d(f_N, f) < \frac \epsilon 3 + d(f_N,f) < \frac \epsilon 3 + \frac {2\epsilon}{3} = \epsilon$$
Thoughts: it seems the confusion in your question comes in Step 3 -- I hope I have made clear to you which bounds come from the uniform $d(f,g)$ type bounds and which bounds must be argued pointwise.
