1. Prove that the set of $p$-integrable functions on $[0,1]$ with the Lebesgue measure $\lambda$ is separable using Lusin's Theorem and the Stone Weierstrass Theorem.
2. Prove that $\mathcal{L}_\infty$ is not separable by constructing a set $\{x_\alpha : \alpha \in [0,1] \}\subset \mathcal{L}_\infty$ such that $||x_\alpha -x_\beta || =1$ if $\alpha\neq \beta$.
Attempt at 1: Given $f\in \mathcal{L}_p$ I know from Lusin's theorem that there is an open set $E_f$ such that $\lambda(E_f)<\epsilon$ and $f|_{[0,1]\backslash E_f}$ is continuous. Since $[0,1]\backslash E_f$ is compact, from Stone-Weierstrass there is a polynomial $p_f$ defined on $[0,1]\backslash E_f$ (which we can safely assume has rational coefficients) such that $\sup_{[0,1]\backslash E_f}{|f(x)-p(x)|}<\epsilon.$
Now we can extend the polynomial $p_f$ to all of $[0,1]$ by making it $0$ in $E_f$, and this new polynomial is in $\mathcal{L}_p$ since it is bounded in $[0,1]$.
Now: $$\|x-p\|^p_p=\int|f-p|^p d\lambda=\int_{[0,1]\backslash E_f}|f-p|^p d\lambda+\int_{E_f}|f|^p d\lambda.$$
I (believe) can make both terms on the right-hand side as small as desired. The problem is the that the polynomials with rational coefficients in a given set $E_f$ form a countable set, but for each $E_f$ I have a different polynomial, and I've got as many $E_f$ (and polynomials) as elements of $\mathcal{L}_p$. How can I make this countable?
Also, are there any mistakes in my attempt?
As to question number 2, I'm completely stumped and could use any help.
Thanks in advance.