Show that $f_n \to f$ over $\| \cdot \|_\infty$ implies $f_n \to f$ over $\| \cdot \|_2$ and is $(C[a,b], \|\cdot\|_2)$ Banach?

Part (i) looks fine, for the second part, define the sequence, $$ f_n(x)= \begin{cases} 0,& a\leq x\leq a+\frac{b-a}{2}\\ \frac{2n}{(b-a)}(x-a-\frac{b-a}{2}),& a+\frac{b-a}{2}<x\leq a+ \frac{b-a}{2}+\frac{b-a}{2n}\\ 1,& a+\frac{b-a}{2}+\frac{b-a}{2n}<x\leq b \end{cases} $$ for which it is definitely better to draw a picture (we are taking continuous ramps between $0$ and $1$, with steeper slope, giving the ramp less and less time to get to $1$ as $n$ gets large). The point is that $f_n(x)$ is approximating the discontinuous function $\chi_{[\frac{b-a}{2},1]}(x)$ in $||\cdot||_2$, since $$ ||f_n(x)-\chi_{[\frac{b-a}{2},1]}(x)||\leq \frac{b-a}{4n} $$ where the bound comes from the area of the triangle between the two graphs (everything is $\leq 1$, so squaring only helps us). Since each $f_n$ is continuous and has a discontinuous limit in this metric, $(C([a,b]),||\cdot||_2)$ is not a Banach space.