For subsets of $S_1$ and $S_2$ of $\mathbb{R}^n$, you can write $S_1+S_2$ as a union of translates. Namely,
Given $X\subseteq \mathbb{R}^n$ and $\mathbf{a}\in\mathbb{R}^n$, $X+\mathbf{a} = \{ \mathbf{x}+\mathbf{a}\mid \mathbf{x}\in X.\}$ It is not hard to show that if $X$ is open, then so is $\mathbf{X}+\mathbf{a}$; and that if $X$ is closed then so is $X+\mathbf{a}$.
Given subsets $X,Y\subseteq \mathbb{R}^n$, then $\begin{align*} X+Y &= \{\mathbf{x}+\mathbf{y}\mid \mathbf{x}\in X,\mathbf{y}\in Y\}\\ &= \bigcup_{\mathbf{y}\in Y} (X+\mathbf{y})\\ &=\bigcup_{\mathbf{x}\in X} (Y+\mathbf{x}). \end{align*}$
So here, $S_1$ and $S_2$ are each closed; their sum is an infinite union of closed sets, and thus may or may not be closed a priori (it's not closed in fact).
(An infinite union of closed sets can easily be open: take for example $\bigcup_{n=3}^{\infty} \left[\frac{1}{n},1-\frac{1}{n}\right] = (0,1),$ which is an open set expressed as a countable union of closed intervals).
However, $S_1+S_2$ in this case happens to be open. The reason it is not clopen is because it is not closed.
As for $S_1\cup S_2$, it's a union of closed sets, so it is closed (which boundary points do you believe it does not contain?); but it is not clopen (because it is not open). The point $(1,1)$ is in $S_1$, but no neighborhood of $(1,1)$ is completely contained in $S_1\cup S_2$.
(In fact, $\mathbb{R}^n$ does not have any clopen sets except for $\emptyset$ and $\mathbb{R}^n$; that's because $\mathbb{R}^n$ is connected).