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 $B\cap U\cap V = \emptyset$.
to show that if $X$ is a compact Hausdorff space then $x,y$ are in the same quasicomponent if and only if they are in the same component.
This is from "General topology" by Riszard Engelking.
It seems the following.
to show that the minimal element is connected
Let $A$ be a minimal element of the family $\mathcal{A}$. Suppose that the set $A$ is not connected. Then $A$ is a union of two its disjoint clopen non-empty subsets . Since $x$ and $y$ are in the same quasicomponent of $A$, the set $\{x,y\}$ is contained in one of these two sets. Denote this set of $B$. Since the set $B$ is a clopen subset of $A$, each quasicomponent of the set $B$ is a quasicomponent of the set $A$. Hence $B\in\mathcal A$. But $B$ is a proper subset of the set $A$, which contradicts to the minimality of the set $A$.
Best Answer
It seems the following.
Fix any element $K\in\mathfrak{C}$. Then the set $C=\bigcap \mathfrak{C}$ is compact as an intersection $\bigcap\{ K\cap L:L\in\mathfrak{C}\}$ of a family $ \{ K\cap L:L\in\mathfrak{C}\}$ of closed subsets of a compact set $K$.
Assume that the set $C$ is not connected. This means that $C$ can be represented as a union $C_1\cup C_2$ of two its disjoint non-empty clopen (that is, closed and open) subsets. Since $C_1$ and $C_2$ are disjoint compact subsets of a Hausdorff space $X$, it is well known and easy to prove that they can be separated by disjoint open neighborhoods, that is there exist two disjoint open subsets $U_1$ and $U_2$ of the space $X$ such that $C_1\subset U_1$ and $C_2\subset U_2$. Put $U=U_1\cup U_2$.
Now we are going to find a set $L\in\frak C$ such that $L\subset U$. If $K\subset U$, we are done. In the opposite case consider a family ${\frak C}’=\{(K\cap L)\setminus U:L\in\frak C\}$. Since $\bigcap{\frak C}=C\subset U$, $\bigcap \mathfrak{C}'=\varnothing$. Since $\frak C’$ is a family of compact subsets of the compact space $K$, it contains an empty element $(K\cap L)\setminus U$. Since the family $\frak C $ is a chain, this implies that there exists a set $L\in\frak C$ such that $L\subset U$. Then $(L\cap U_1)\cup (L\cap U_2)$ is a partition of the set $L$ into two its disjoint clopen subsets. Each of these sets is non-empty. Indeed, since $\frak C$ is a chain, $L\supset \bigcap \mathfrak{C}=C$. Then $L\cap U_i\supset C\cap U_i\ne\varnothing$. So the set $L$ is not connected, a contradiction.