Prove $C^1[a,b]$ is a Banach space. 
Let $C^1[a,b]$ be the space of continuous differentiable functions on $[a,b]$ equipped with the following norm 
  $$\|x\|=|x(a)| + \sup_{t\in [a,b]}|x'(t)|.$$
  Prove that $(C^1[a, b],\|\cdot\|)$ is a Banach space.

Here is my attempt:
Let $(x_n)$ be a Cauchy sequence in $C^1[a,b]$, i.e., $$\lim \limits_{m,n\to \infty}\|x_m-x_n\|=0.$$ 
This means
$$|x_m(a) -x_n(a)| + \sup_{t\in [a,b]}|x_m'(t) -x_n'(t)| \to 0,$$
as $m,n\to\infty$. Therefore, $\{x_n(a)\}$ is a Cauchy sequence in $ \mathbb R$ and $\{x'_n\}$ is a Cauchy sequence in $C[a,b]$ with $\sup$ norm. Thus $\{x_n\}$ converges at $a$, and $\{x'_n\}$ uniformly converges to some $y$ in $C[a,b]$. 
I don't know how to show that $y\in C^1[a,b]$ and $x_m \to y$ in $C^1[a,b]$.

In your notation, let $c=\text{lim}_{n\rightarrow \infty}x_n(a)$: Define
$$
Y(x)=\int_a^x y(t)dt+c.
$$
Then by the fundamental theorem of calculus $Y\in C^1(a,b)$, $Y'(x)=y(x)$ and furthermore $Y(a)=c$. Now show that $Y$ is the limit of $x_n$
