Verify a Subset of $C[0,1]$ that is also closed as a subset of $L^2[0,1]$ Is Finite Dim

Question

I'm working on the following question in Royden:

Let $X$ be a linear subspace of $C[0,1]$ that is closed as a subset of $L^2[0,1]$. Verify the following assertions to show that $X$ has finite dimension. Suppose the sequence $(f_n)$ belongs to $X$.

1. Show that $X$ is a closed subspace of $C[0,1]$

2. Show that there is a constance $M \geq 0$ such that for all $f \in X$ we have $\|f\|_2\leq \|f\|_{\infty}$ and $\|f\|_{\infty} \leq M\|f\|_2$

3. Show that for each $y\in [0,1]$, there is a function $k_y$ in $L^2$ such that for each $f \in X$ we have $f(y) = \int_0^1k_y(x)f(x) dx$.

My Attempts and Questions:

1. Assuming $C[0,1]$ has the sup norm, then any sequence in a linear subspace will, in fact have a limit in $C[0,1]$, because continuous functions over a compact set are uniformly continuous and a limit of such functions is also uniformly continuous in the sup metric.

2. $\|f\|_2\leq \|f\|_{\infty}$ amounts to pulling the max out of the integral. The other direction I adopted this argument.

3. I'm not sure how to tackle this. Any ideas?