Disjoint convex sets which cannot be separated by any continuous linear functional This problem is out of Rudin's Functional analysis exercise 3.2. The problem is stated below. I'm really struggling with this chapter in general. It has a lot of new topics I have not seen before. Any help is appreciated. I am attaching some of my current thoughts on the problem. 
Suppose $L^2 = L^2([-1,1])$, with respect to Lebesgue measure. For each scalar $\alpha$, let $E_\alpha$, be the set of all continuous functions $f$ on $[-1,1]$ such that $ f(0) = \alpha$. Show that $E_\alpha$ is convex and that each is dense in $L^2$. Thus $E_\alpha$ and $E_\beta$ are disjoint convex sets (if $\alpha \neq \beta$) which cannot be seperated by any continusous linear functionals $\Lambda$ on $L^2$. Hint: What is $\Lambda(E_\alpha)$. 
proof:
Suppose the given hypothesis above. 
To show convexity let functions $f, g \in E_\alpha$ be arbitrary. Then we observe that $h(x) = tf(x) + (1-t) g(x)$ for $0 \leq t \leq 1$ is a continuous function and that $h(0) = t f(0) + (1-t)g(0) = t\alpha + (1-t)\alpha = \alpha.$ Therefore $h(x) \in E_\alpha$ and we have $E_\alpha$ convex. 
To show that $E_\alpha$ is dense in $L^2$. Let $A$ be a collection of continuous functions in $L^2$. Need to show that $E_\alpha \cap A \neq \emptyset$, but not sure how... I assume that given $\epsilon > 0$ we need to take an arbitrary function $f \in A$ and show there exists a function $g \in E_\alpha$ such that for $h(x) = f(x) - g(x)$,
$$
\|h(x)\| = \left ( \int_{-1}^{1}|h(x)|^2 d\mu \right )^{1/2} < \epsilon
$$
It makes sense that the sets $E_\alpha$ and $E_\beta$ are disjoint because every function in each of them will have a different value at $0$. But the last statement confuses me. 
Also I am struggling with the hint, is $\Lambda(E_\alpha) = 0$? 
I know that $L^2$ is its own dual space, since $p=2$ and its conjugate $q=2$ as well, so $\Lambda \in L^2$. It seems like symmetry of interval may have something to do with it, along with convexity or $E_\alpha$. 
Thanks again for any help in advance. 
 A: $X=L^2[-1,1]$ is the completion of the continuous functions on $[-1,1]$ using the
$L^2[-1,1]$ norm.
Let $\epsilon>0$ $f \in X$ and choose $c$ such that $\|f-c\|_2 < {1 \over 2} \epsilon$.
Let $M=\max (|\alpha|, \sup_x |c(x)|)$ and choose $\delta>0$ such that $2 \delta M^2 < {1 \over 2} \epsilon$.
Define $c^*(x) = c(x)$ for $|x|> \delta$, and straight line interpolation
between the three points $(-\delta, c(-\delta)), (0, \alpha), (\delta, c(\delta))$. It is straightforward to see that $\|c-c^*\|_2 \le 2 \delta M^2$,
and so $\|f-c^*\|_2 < \epsilon$ and $c^*(0) = \alpha$. Hence $E_\alpha$ is dense
in $X$ for all $\alpha$.
Choose a non zero $\Lambda \in X^*$. Then $\Lambda(f) \neq 0$ for some $f$ and since
$\Lambda( c f) = c \Lambda (f)$ for all scalars $c$, we see that
$\Lambda(X)$ is the entire set of scalars. Since $E_\alpha$ is dense in $X$, we see that $\Lambda(E_\alpha)$ is dense in the scalar field.
It follows that $E_\alpha \neq E_\beta$ (with $\alpha \neq \beta$) 
cannot be separated.  
