I do this proof in 3 steps
First I pick a Cauchy sequence of $f_n(x) \in C[a,b]$ and then fix an $\epsilon >0$ and some $x \in [a,b]$ then constitute $f(x)$ s.t $\lim_{n \to \infty} f_n(x)=f(x)$ in absolute value.
then I try to prove that this limit $f(x)$ is continuous and $\lim_{n \to \infty} f_n(x)=f(x)$ in sup-norm.
Which is a very long proof.
The second proof is.
Let $\{f_n\}$ be a Cauchy sequence in $C[a,b]$ then for all $\epsilon >0$ there exists an $N(\epsilon)$ s.t. for all $n,m>N(\epsilon)$ we have $\| f_m -f_n \|_{\infty} < \epsilon$ which implies $| f_m(x) - f_n(x) | <\epsilon$ for all $x \in [a,b]$ then $\{f_n(x)\}$ is uniformly cauchy on $[a,b]$. Then by the following Theorem.
Theorem. A sequence $\{f_n\}$ of functions $f_n : A \to\mathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$.
Then we deduce that the $f_n \to f$ uniformly on $[a,b]$
But given that $f_n$ are continuous functions then by uniform convergence we know that $f$ is continuous on $[a,b]$
then by definition of uniform convergence we have that for all $x \in [a,b]$, for all $\epsilon >0$ $\exists N(\epsilon )$ st $\forall n>N(\epsilon )$ $| f_n(x) - f(x) | < \epsilon$ $\implies$ $\| f_n - f \|_{\infty} < \epsilon$
Does the second proof suffice?
In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that $\{f_n\}$ is Cauchy means that given $\epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <\epsilon$ for all $x$ for all $n,m >k$. $\cdots (1)$. In particular, you can fix $x$ an conclude that the sequence $\{f_n(x)\}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m \to \infty$ you get $|f_n(x)-f(x)| \leq \epsilon$ for all $x$ for all $n >k$. This means $f_n \to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.