You have to suppose also that each $S_i$ is non empty.
We prove the more general :
Theorem The intersection $S$ of any decreasing sequence $(S_i)_{i\ge0}$ of nonempty compact and connected subsets of some metric space $(E,d)$ is connected.
(whose proof can be read while thinking at $E$ as beeing $\mathbb{R}^2$).
Proof Let us prove first that, for any open set $\Omega$ containing $S$, there exists an integer $i$ such that $S_i\subset\Omega$ (which implies that the inclusion is also true for any index $>i$).
Suppose the contrary ...
Then, there exists une sequence $(x_i)_{i\ge0}$ such that $x_i\in S_i-\Omega$ for all $i$.
All the $x_i$ belong to $S_0$, which is compact, and therefore we can extract a sequence $(x_{i_p})_{p\in\mathbb{N}}$ which converges to some $y\in E$.
For any $N\in\mathbb{N}$, the truncated sequence $(x_{i_p})_{p\ge N}$ has all its terms in $S_{i_N}$ thus in $S_N$. So $y\in S_N$, because $S_N$ is closed. This proves that $y\in S$.
But $E-\Omega$ is closed in $E$, and therefore $y\not\in\Omega$, a contradiction.
Now, suppose $S=(S\cap \Omega_1)\cup(S\cap\Omega_2)$, where $\Omega_1$, $\Omega_2$ are disjoint subsets of $E$, such that $S\cap\Omega_1\neq\emptyset$ and $S\cap\Omega_2\neq\emptyset$.
Put $\Omega=\Omega_1\cup\Omega_2$. We can see that $\Omega$ is an open set, which contains $S_i$ for some $i\in\mathbb{N}$. But $S_i$ beeing connected, there exists $\alpha\in\{1,2\}$ such that $S_i\cap\Omega_\alpha=\emptyset$. And so $S\cap\Omega_\alpha=\emptyset$ for the same $\alpha$. Contradiction !
