For $C^1[a,b]$, define $||f||= ||f||_{sup}+||f'||_{sup}$. Prove that the norm is in fact a norm and that $C^1[a,b]$ is a complete normed linear space

Question

A norm has to fit three properties.

1. $\||f||= ||f||_{sup} + ||f'||_{sup}$
2. $\||af||=|a|||f||$
3. $\||f+g|| \leq ||f||+||g||$

How I proved that $||.||$ satisfies these properties:

1. Since $||f||_{sup}$ and $||f'||_{sup}$ are both norms, then $||f||_{sup} \geq 0$ and $||f'||_{sup} \geq 0$.

Then we get $||f||_{sup}+||f'||_{sup}\geq0$

2. $||af||=||af||_{sup}+||af'||_{sup}$

$=|a|||f||_{sup}+|a|||f'||_{sup}$

$=|a|(||f||_{sup}+||f'||_{sup})$

3. Here is where I'm having some trouble. I was something along the lines of

We know $||f+f'||_{sup} \leq ||f||_{sup}+||f'||_{sup}=||f||$, let $||g||=||f||_{sup}$

Then $||f+f'||_{sup}+||g|| \leq ||f||+||g||$

$||f+f'+f||_{sup} \leq ||f+g||+||g|| \leq ||f||+||g||+||g||$

I know this isn't exactly correct but I'm not sure how to approach it.

For the second part of this proof,

Let each $f_n,f'_n \in (C[a,b], ||.||_{sup})$. We know that $C([a,b])$ is complete, thus $ \exists f,g \in C([a,b])$ s.t. $f_n \rightarrow g$ wrt $||.||_{sup}$. Let $F_n(x)=\int_{a}^{x}f_n{t}dt, F(x)=\int_{a}^{x}f(t)dt$, then $F'_n \rightarrow F$ uniformly because $||F_n-F||_{sup} \leq sup \int_{a}^{x} |f_n(t)-f(t)|dt \leq ||f_n-f||_{sup}< \epsilon$ for $a \leq x \leq b$.

From FTC, $f_n(x)-f_n(a)+\int_{a}^{x}f_n'(t)dt$

Since $f_n'\rightarrow g$ uniformly, $f(x)-f(a)=\int_{a}^{x}g(t)dt$ where $f'=g$.

Since $f_n \rightarrow f$ and $f_n' \rightarrow g=f'$, $f_n \rightarrow f \in C([a,b])$ wrt to $||.||$.

I'm not sure if this proof holds.