1. limit points

Prove that a set $\displaystyle S \subset \bold{R}$ is closed if and only if it contains all of its limit points.

Suppose that $\displaystyle S$ is closed, and $\displaystyle x_0$ is a limit point of $\displaystyle S$. Then $\displaystyle S^{C}$ is open so that there is an $\displaystyle \varepsilon > 0$ such that $\displaystyle (x_{0} - \varepsilon, x_{0} + \varepsilon) \cap S = \emptyset$ which is a contradiction.

If $\displaystyle S$ contains all its limit points, then there exists an $\displaystyle \varepsilon > 0$ such that $\displaystyle (x_{0}- \varepsilon, x_{0} + \varepsilon) \cap S \neq \emptyset$. So $\displaystyle S^{C}$ is open $\displaystyle \implies S$ is closed.

Is this correct?

2. You have the correct ideas and concepts but an awful form.
The first part should go something like this. Say that $\displaystyle x_0$ is limit point of of the closed set $\displaystyle S$. Suppose $\displaystyle x_0 \notin S$ then $\displaystyle x_0 \in S^c$ which is an open set. So $\displaystyle \left( {\exists \delta > 0} \right)\left[ {\left( {x_0 - \delta ,x_0 + \delta } \right) \subseteq S^c } \right]$. But that is a contradiction to $\displaystyle x_0$ being a limit point.

No you try to rewrite the second part.

3. $\displaystyle S$ contains all of its limit points. Suppose that $\displaystyle x_0 \not \in S$. Then $\displaystyle x_{0} \in S^{c}$, then $\displaystyle (\exists \delta >0)[(x_0- \delta, x_0+\delta) \subseteq S^{c}]$. Thus $\displaystyle S^{c}$ is open which implies that $\displaystyle S$ is closed.

4. That looks good.
You might add that $\displaystyle x_0 \in S^c$ means that $\displaystyle x_0$ is not a limit point of $\displaystyle S$.