Intersection of Connected Sets This is an old exam question that I don't have a solution to:

Let $X$, a compact Hausdorff (T2) space, and let $\phi$ a family of closed, non-empty, and connected subsets of $X$, such that for every $A, B \in \phi$, $A \subset B$ or $B \subset A$.

Prove that $Y:= \cap \{A: A \in \phi\}$ is connected. 

I tried to solve this question with a friend, and this is what we came up with:


Obviously, $X$ is normal space (T4).
We tried to see what happens if $Y = U_1 \cup U_2$, disjoint and open sets (and?)
We tried to use nets and maybe see if we can create a net that converges to $x \neq y$ (and contradict X being Hausdorff space).
We tried to work with continuous functions, but this idea didn't lead us anywhere either.


I feel like there is a simple observation that we are missing. Any ideas?

Thanks!
Re-corrected: $\newcommand{\cl}{\operatorname{cl}}$Suppose that $Y=H\cup K$, where $H\cap K=\varnothing$, $H\ne\varnothing\ne K$, and $H$ and $K$ are clopen in $Y$. Let $f:Y\to\{0,1\}$ take $H$ to $0$ and $K$ to $1$, use the Tietze extension theorem to extend $f$ continuously to $\hat f:X\to[0,1]$, and let

$$\begin{align*}
U&amp;=\hat f^{-1}\left[\left[0,\frac12\right)\right]\supseteq H\;,\\
V&amp;=\hat f^{-1}\left[\left(\frac12,1\right]\right]\supseteq K\;,\text{ and}\\
F&amp;=\hat f^{-1}\left[\left\{\frac12\right\}\right]\;.
\end{align*}$$

For each $A\in\phi$ we must have $F\cap A\ne\varnothing$, as otherwise $U\cap A$ and $V\cap A$ would be a disconnection of $A$. Thus, $\mathscr{C}=\{F\cap A:A\in\phi\}$ is a nested family of non-empty compact sets. But then $\varnothing\ne\bigcap\mathscr{C}=F\cap Y=\varnothing$, which is absurd.
Seems to me that if either $A \subset B$ or $B \subset A$ 
then for any $ Y = \bigcap [ A \mid A \in \phi ]$ there exists an set in this union, call it $H$, such that for any other set in the union, lets call them $K_\lambda$, $ H \subset K_\lambda$. From which we can conclude that $ Y = H$ and is thusly connected.

Intuitively, there is a minimal element in this union that is induced by the ordering on $\phi$ so this element is our 'limiting factor' so to speak.