Checking if a set is closed 
I am unable to find an example where the point belongs to $S = S_1 + S_2$ but it's not an adherent point to $S$ or vice versa.
An adherent point of a subset $A$ of a topological space $X$, is a point $x \in X$ such that every open set containing $x$ contains at least one point of $A$. 
A set is closed iff it contains all its adherent points.
 A: *

*Show that $(0,0) \notin S_1+S_2$

*For $n \in \mathbb N$ we have $(n,1/n) \in S_1$ and $(-n,1/n) \in S_2$.
Hence the sequence $((0,2/n))$ is a sequence in $S_1+S_2$, but the limit of this sequence is not in $S_1+S_2$
A: To show that a set is not closed you must show that it does not contain all of its limit points.  That is, if you can demonstrate one point which is a limit point of $S_1+S+2$, but is not an element of $S_1+S+2$, then you have proven that $S_1+S+2$ is not closed.
Consider the point $(0,0)$. 
$(0,0) \not\in S_1+S+2$: 
If  $(0,0)$ were in $S_1+S+2$, then there would be some 
$x_1,x_2,y_1,y_2)$ such that 
$$
\left\{ \begin{array}{c}y_1 = \frac1{x_1} \\y_2 = -\frac1{x_2} \\
x_1+x_2 = 0 \\ y_1+y_2 = 0\end{array}\right.
$$
This gives
$$
\left\{ \begin{array}{c}
x_1+x_2 = 0 \\ \frac1{x_1}  -\frac1{x_2}= 0\end{array}\right.
$$
So $x_2=-x_1$ which gives $\frac1{x_1}  +\frac1{x_1} = 0$, which is not true for any $x_1\in\Bbb R$.
$(0,0)$ is a limit point of $S_1+S+2$:
Let $\epsilon \in \Bbb R$ with $\epsilon > 0$.  Look at the equations
$$
\left\{ \begin{array}{c}y_1 = \frac1{x_1} \\y_2 = -\frac1{x_2} \\
x_1+x_2 = 0 \\ y_1+y_2 = \epsilon\end{array}\right.
$$
For this set of equations, 
$x_2=-x_1$ which gives $\frac1{x_1}  +\frac1{x_1} = \epsilon$, which has the solution $x_1 = \frac2\epsilon, x_2 = -\frac2\epsilon, y_1 = y_2=\frac \epsilon{2}) $
Thus $(0,\epsilon) \in S_1+S_2$. Since this holds for arbitrary positive $\epsilon$, the sequence of points corresponding to $\epsilon = 2^{-n}$ demonstrates that $(0,0)$ is a limit point of $S_1+S_2$.  
