My professor gave a proof of the completeness of $(C^1(\bar \Omega),\|\cdot \|_{C^1})$ based on the fundamental theorem of calculus. I though about an alternative and I would like to know whether this is formally correct:
Assuming completeness of $(C^0(\bar \Omega),\|\cdot \|_{C^0})$ is known, we want to show that, given a Cauchy sequence $(f_n)_n \subset C^1(\bar \Omega)$ there is a function $f \in C^1(\bar \Omega)$ s.t. $f_n \to f$ in the $C^1$-norm, i.e. $ \| f_n - f \|_{C^1} \to 0$.
Knowing that
$$ \| f \|_{C^1} \triangleq \| f \|_{C^0} + \| f' \|_{C^0}$$
where $f'$ is the first derivative of $f$. It is immediate to show that
$$ (f_n)_n \text{ is Cauchy in } C^1(\bar \Omega) \Longrightarrow (f_n)_n,(f_n')_n \text{ are Cauchy in } C^0(\bar \Omega)$$
By completeness of $C^0(\bar \Omega)$ there are $f,g \in C^0(\bar \Omega)$ s.t. $f_n \xrightarrow{C^0} f, ~ f_n' \xrightarrow{C^0} g$. To conclude we need to show that $f'=g$. I would now argue as:
$$ \sup_{x \in \bar \Omega} \bigg|\lim_{h \to 0} \bigg(\frac{f(x+h) - f(x)}{h} -\lim_{n \to \infty} \frac{f_n(x+h) - f_n(x)}{h} \bigg) \bigg| = \\ \sup_{x \in \bar \Omega}~ \lim_{h \to 0}~ \lim_{n \to \infty} \frac{1}{h} \bigg| f(x+h) - f(x)- f_n(x+h) + f_n(x)\bigg| \le \\ \sup_{x \in \bar \Omega}~ \lim_{h \to 0}~ \lim_{n \to \infty} \frac{1}{h} \bigg| f(x+h) - f_n(x+h)\bigg| + \bigg|f_n(x)- f(x)\bigg| = 0$$
as, by hyp
$$ \sup_{x \in \bar \Omega} |f(x) - f_n(x)| \xrightarrow{n \to \infty} 0$$
so this holds in particular for any $x \in \bar \Omega$.
My doubts are about the last claim, where I conclude that $\sup (…) = 0$, I fear I am messing with $\sup \lim$ vs $\lim \sup$ and that the last step is not legitimate.
It doesn't matter whether the last step is legitimate or not, because you're simply not proving what needs to be proved. As you point out, you need to show that $f'=g$. But there's no $g$ anywhere in your supposed proof of this!
There's no $g$ visible. There's also no implicit hidden $g$. By definition $g=\lim_nf_n'$, and there's also no $f_n'$ in the proof. No, there's no $f_n'$ in the display $$\lim_{h\to0}\lim_{n\to\infty}\frac{f_n(x+h)-f_n(x)}{h}.$$That's the derivative of $\lim_n f_n$, not the derivative of $f_n$.
In fact all you've proved here is this: $$\lim_{h\to0}0=0.$$Because the thing on the inside of the $\lim_h$ on the first line is $$\frac{f(x+h)-f(x)}{h}-\lim_{n\to\infty}\frac{f_n(x+h)-f_n(x)}{h}=0.$$So you're taking the limit of $0$ as $h$ tends to $0$. Doesn't prove anything about $f'$.
How do you prove $f'=g$? The only thing that springs to mind is an argument using the Fundamental Theorem of Calculus...