I think the following is a counterexample: take $Y=[-1,1]\times\{a,b\}$, where $[-1,1]$ has the standard topology, and $\{a,b\}$ the discrete topology, and let $X=Y/\sim$, $y\sim y'$ if and only if there exist $x\in[0,1]$ such that $y=(x,a)$ and $y'=(x,b)$, or $y=(x,b)$ and $y'=(x,a)$. That is, identify all points except $0$. Give $X$ the quotient topology
This is the interval $[-1,1]$ with "a doubled origin", a common proving ground because the space is $T_1$ but not $T_2$, but any two points other than the doubled origins can be separated by open neighborhoods. (So, in a sense, it is "almost" Hausdorff; the Hausdorff property only fails for one choice of points, and there are lots of other points around).
Since $Y$ is compact and the quotient map is continuous and onto, $X$ is compact.
For every positive integer $n$, let $\mathcal{B}_n\subseteq Y$ be the set $[-\frac{1}{n},\frac{1}{n}]\times\{a,b\}$, and let $B_n$ be the image of $\mathcal{B_n}$ in $X$; that is, $B_n$ is the interval from $-\frac{1}{n}$ to $\frac{1}{n}$, including both origins.
$B_n$ is closed, since $X-B_n = [-1,-\frac{1}{n})\cup(\frac{1}{n},1]$ is a union of two open sets. It is also connected, because $B_n$ is a union of two connected subsets (the two copies of the interval $[-\frac{1}{n},\frac{1}{n}]$ obtained by removing one of the two $0$s) and the two subsets intersect.
What is $\cap_{n=1}^{\infty}B_n$? It's a set whose only two elements are the doubled origin points. But this subset of $X$ is not connected, because $X$ is $T_1$, so there exist open neighborhoods $U$ and $V$ such that $(0,a)\in U-V$ and $(0,b)\in V-U$. So $B\subseteq U\cup V$, $U\cap B\neq\emptyset \neq V\cap B$, and $B\cap U\cap V = \emptyset$.
The problem is not with your understanding of divided, but rather with your understanding of closed. In the space $X=(0,1)\cup(2,3)$, the sets $(0,1)$ and $(2,3)$ are closed. This is because the topology $\tau$ on $X$ is the subspace (or relative) topology inherited from $\Bbb R$. A subset $U$ of $X$ is open in $X$ if and only if there is a $V\subseteq\Bbb R$ such that $V$ is open in $\Bbb R$ and $V\cap X=U$. Of course $(0,1)$ is open in $\Bbb R$, and $(0,1)\cap X=(0,1)$, so $(0,1)$ is open in $X$. By the definition of closed set this means that $X\setminus(0,1)$ is closed in $X$. And $X\setminus(0,1)=(2,3)$, so $(2,3)$ is closed in $X$. A similar argument shows that $(0,1)$ is also closed in $X$. Indeed, both of these sets are clopen (closed and open) as subsets of $X$, even though they are only open as subsets of $\Bbb R$. Openness and closedness depend not just on the set, but on the space in which it is considered.
You have the same problem with your first example: the sets $[0,1]$ and $[2,3]$ are clopen in the subspace $Y=[0,1]\cup[2,3]$ of $\Bbb R$, not just closed. For example, $[0,1]=\left(-\frac12,\frac32\right)\cap Y$, and $\left(-\frac12,\frac32\right)$ is open in $\Bbb R$, so $[0,1]$ is open in $Y$.
Best Answer
i) Not true, as you said, though union of two sets with one point in common is connected.
ii) Not true. Take two cheetos reflected.
iii) True. You can use the characterization that $X$ is connected iff every continuous function from $X$ to the discrete space $\{0,1\}$ is constant. Or you can simply find a smart way of showing that $X \times Y$ is the union of pairwise non-disjoint connected spaces.