1. ## limit points

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

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

If $S$ contains all its limit points, then there exists an $\varepsilon > 0$ such that $(x_{0}- \varepsilon, x_{0} + \varepsilon) \cap S \neq \emptyset$. So $S^{C}$ is open $\implies S$ 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 $x_0$ is limit point of of the closed set $S$. Suppose $x_0 \notin S$ then $x_0 \in S^c$ which is an open set. So $\left( {\exists \delta > 0} \right)\left[ {\left( {x_0 - \delta ,x_0 + \delta } \right) \subseteq S^c } \right]$. But that is a contradiction to $x_0$ being a limit point.

No you try to rewrite the second part.

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

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