$C^1[0, 1]$ is complete with the $C_1$ norm. Consider the following norm, $$||f||=|f(0)|+||f^\prime||_\infty$$ This norm is equivalent to $C_1$ norm. But in the book they asked to prove the space is complete using fundamental theorem of calculus and convergence theorem of Riemann integration.
How to prove the result, without using the fact that the norms are equivalent?
Let $(f_n)$ be a Cauchy sequence of $(C^1, ||.||)$, then the sequence of derivatives $(f_n')$ is a Cauchy sequence in $(C^0,||.||_{\infty})$. This last space is complete so $(f_n')$ converges to a function $g$ that is continuous over $[0,1]$. Now we only need to show that $(f_n)$ converges to an antiderivative of $g$ to have that $(f_n)$ converges in $C^1$. Using again the Cauchy property satisfied by $(f_n)$, $(f_n(0))$ is a Cauchy sequence of $\mathbb{R}$ that is complete and thus this last sequence converges to a real number $a$. Now there is no choice for the candidate limit of $(f_n)$, it is an antiderivative of $g$ taking the value $a$ at $0$. We introduce $$ f : x \longmapsto a + \int_0^xg(t)dt $$ that is a $C^1$ function over $[0,1]$ by the fundamental theorem of calculus, let us check that $(f_n)$ goes to $f$, $$ ||f_n-f|| = |f_n(0)-f(0)| + ||f_n'-f'||_{\infty} = |f_n(0)-a| + ||f_n'-g||_{\infty} $$
where the two last terms converge to $0$ by construction.