Proving that the set of limit points of a set is closed

Your argument is correct, but incomplete: All you need to finish it is to ensure that you can find a neighborhood of $y$ contained in the neighborhood of $x$ that you began with (any will do, since all contain elements of $E$). Use the triangle inequality to find an appropriate radius for the neighborhood of $y$.

Let $\hat S$ be the set of all limit points of $S$. Prove that $\hat S$ is a closed set.

Proof: Suppose $x_0$ is a limit point of $\hat S$. Then given $\varepsilon > 0$ there exists $x \in \hat S$ with $\vert x - x_0\vert < \frac\varepsilon2$. Now $x \in \hat S$ is a limit point of $S$ so there exists $x' \in S$ such that $\vert x' - x \vert < \frac\varepsilon2$. Now $$\vert x' - x_0 \vert = \vert x' - x + x - x_0 \vert \leq \vert x' - x \vert + \vert x - x_0 \vert < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$$ Thus $x_0$ is a limit point of $S$ and by definition is contained in $\hat S$. We have shown that $\hat S$ contains all of its limit points. By theorem that states that a set is closed if and only if it contains all its limit points, we have just shown that $\hat S$ is a closed set.

Let $x$ be a limit point of $E'$, and let $\varepsilon >0$. Then (by definition) there exists $y\in E'$ such that $0<d(x,y)<\frac{\varepsilon}{2}$. Since $y\in E'$ there exists $z\in E$ such that $0<d(y,z)<d(x,y)$ (here one uses $d(x,y)$ as the epsilon from the definition of a limit point).

By triangle inequality we have $d(x,z)\leq d(x,y)+d(y,z)<\varepsilon$, and note that indeed $x\neq y$. So (by definition) $x$ is a limit point of $E$. That is $E'' \subseteq E'$, which proves that $E'$ is closed.