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Since $C[a,b]$ and $C[0,1]$ are bi-lipschitz homeomorphic for every $a,b$, one's completeness is equivalent to the other, so we can prove it only for $C[0,1]$.
I found a few answers (like this or this), but I believe one detail is lacking.

Let's use e.g. $$f_n(x)=\begin{cases} 1,&\text{if }0\le x\le\frac12\\ 1-n(x-\frac{1}{2}),&\text{if }\frac12\le x\le \frac{1}{2}+\frac{1}{n}\\\\ 0,&\text{if }\frac{1}{2}+\frac{1}{n}\le x\le 1\;. \end{cases}$$ as a counter example. It's a Cauchy series indeed, and assuming it converges to $f\in C[0,1]$:
$\lim_{n\rightarrow\infty}\int_0^1|f_n(x)-f(x)dx=\lim_{n\rightarrow\infty}\Vert f_n-f\Vert=0$
we wish to conclude that
$\forall x\in[0,1]\ f_n(x)-f(x)\rightarrow 0$
thus $f$ has a discontinuity at $x=0.5$, meaning $f\notin C[0,1]$ contradicting our assumption.
However, while $\int|f_n(x)-f(x)\vert dx=0$ (for $f,f_n\in C[0,1])$ does apply $f_n \equiv f$, the limit $\lim_{n\rightarrow\infty}\int|f_n(x)-f(x)\vert dx=0$ doesn't necessarily apply that.
Why then $f(x)=\lim_{n\rightarrow\infty}f_n(x),\ \forall x\in[0,1]$?

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galra
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Assume that $f_n$ indeeds converges in $L^1$ to some $f \in C[0,1]$. Let $x_0 \in [0,\frac{1}{2})$ and assume that $f(x_0) \neq 1$. Then using the continuity of $f$, we can find $\varepsilon,\delta > 0$ such that $|f(x) - 1| > \varepsilon$ for all $x \in [x_0 - \delta,x_0 + \delta]$. But then

$$ 2 \delta \varepsilon \leq \int_{x_0 - \delta}^{x_0 + \delta} |1 - f(x)| \, dx = \int_{x_0 - \delta}^{x_0 + \delta} |f_n(x) - f(x)| \, dx \leq \int_0^1 |f_n(x) - f(x)| \, dx \to 0 $$

a contradiction. This implies that $f(x_0) = 1$ for all $x_0 \in [0,\frac{1}{2})$. Similarly, you can show that if $x_0 \in (\frac{1}{2},1]$ then $f(x_0) = 0$. But since $f$ is continuous, this implies both $f \left( \frac{1}{2} \right) = 0$ and $f \left( \frac{1}{2} \right) = 1$, a contradiction.

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