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
Note that $Q_p$ is an intersection of closed sets, so it is closed, and hence compact, because $X$ is compact. Furthermore, the family which we intersect to get $Q_p$ forms an open cover of $Q_p$, so by compactness it has a finite subcover $A_1,\ldots A_n$. Since the topology is closed under finite intersections, we get that $Q_p$ is also open. Therefore the family of its open subsets in the relative topology $\mathcal{T}_{Q_p}$ is a subset of $\mathcal{T}$, same for closed subsets.
Now assume that there $Q_p$ has a non-trivial clopen subset $C$. Note that it may not contain $p$ by definition of $Q_p$. However, we also get that $Q_p\setminus C$ is clopen in $C$ and since both $\mathcal{T}_{Q_p}\subset\mathcal{T}$ and $\mathcal{F}_{Q_p}\subset \mathcal{F}$, we find that $Q_p\setminus C$ is clopen in $X$ and contains $p$, which contradicts the definition of $Q_p$.