From the fact that $S$ is not closed, you should have concluded that $S$ is not compact.
Connectedness (in fact, path-connectedness) can be verified without explicitly rewriting $S$ in a more comprehensible way (i.e., without recognizing that $S=[0,1)$. Note that $S$ is of the form $S=\{\,f(x)\mid x\in\mathbb R\,\}$ where $f$ is a continuous function. Then between any two points $s_0=f(t_0)$ and $s_1=f(t_1)$ of $S$ we always have a path $[0,1]\to S$, $\tau\mapsto f(t_0+\tau\cdot(t_1-t_0))$.
Of course, once you did identify $S$ as $[0,1)$ you should immediately recall (as you did in your reasoning) that such a half-open interval is not compact and that every interval is connected.
1) We have some fiddley-ness at the start to get $U, V$ disjoint. It's a bit long, so see Disjoint compact sets in a Hausdorff space can be separated
This is possible because the space is Hausdorff (given that it is metrizable). There's a nice phrase by Willard, who says that this is an example of compact sets behaving like points - as $A, B$ are closed in a compact space, they are also compact, and because the space is Hausdorff, there are disjoint open sets containing $A$ and $B$. Neat!
The reason we want such sets will become apparent as we wade on.
2) $U\cup V$ is special because it covers $\bigcap_{i=1}X_i$, and because it's open. So defining $F_i\equiv X_i\setminus (U\cup V)$ is a closed subset of $X$, hence compact, and doesn't contain the intersection $\bigcap_{i=1}X_i$; finally note the $F_i$ also form a nested sequence (draw a diagram!). So if all $F_i$ were non-empty, their intersection would be non-empty - but that's not true!
As $$\bigcap_{i=1}F_i=\bigcap_{i=1}(X_i\setminus (U\cup V))=(\bigcap_{i=1}X_i)\setminus (U\cup V)=\emptyset$$ because we already knew $U\cup V$ covers that intersection.
So, there is some specific $F_k$ that is empty!
3) Nearly there now. As this $F_k$ is defined as $X_k\setminus (U\cup V)$, then if $F_k=\emptyset$, that implies every element of $X_k$ was also in $U\cup V$. i.e. $X_k\subset (U\cup V)$.
4) So we have that there is a pair of disjoint open sets covering this connected $X_k$. But surely they provide a separation of $X_k$? Just taking the complements of the open sets $U, V$ (namely our original $A$ and $B$...) and then intersecting with $X_k$ to get a separation. The only way this wouldn't be true is if $X_k$ had trivial intersection with one of those complements, i.e. if $X_k$ was entirely contained by one of the open sets. But this can't be true, because we know $$X_k\cap A\supseteq (\bigcap_{i=1}X_i)\cap A\neq\emptyset$$
The same holds for $B$, and so $X_k$ really has a separation. This is a contradiction because it was supposed to be connected!
Phew. I hope that makes sense, please ask for clarifications if anything is too terse. As often seems to be the case, a short looking proof is actually an example of economy of writing, rather than a simple idea. That's not to say this isn't a nice proof, it's just not one you can 'see' in a flash without having spent some time thinking topologically.
Best Answer
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 !