Check proof about connectedness Let $X$ topological space and pick $x\in X$. Now write
$$ A= \bigcap_{x\in C_{\lambda}\subset X} C_{\lambda}$$ where the $C_{\lambda}$ are closed, so that $A$ is the smallest closed in $X$ containing $x$. I want to prove that $A$ is connected, so I suppose otherwise and find $C$ and $D$ proper clopen subsets such that $C\cap D=\emptyset$ and $C\cup D=A$. Now $A$ is clearly a closed in $X$ and both $C$ and $D$ are closed in $A$ which in turn makes them closed in $X$. So now one of them must contain $x$ so $A$ is in fact contained in either $C$ or $D$ thus contradicting the assumption that $C$ and $D$ where proper clopens in $A$.
I do not see right now where I am wrong. But in my notes the assumption for this to be true was that $X$ should be compact and $T_2$. And the proof that I haven in my notes is much longer of course and uses both assumptions.
 A: The original statement is not for all closed sets containing $x$.
In any space $\bigcap\{ C: x \in C, C \text{ closed }\}  = \overline{\{x\}}$, and this is connected in any space.
So that would be uninteresting. 
But more interesting: let $X$ be compact and Hausdorff and let $C_i, i \in I$ be downward directed: $$\forall i, j \in I \exists k \in I: C_k \subseteq C_i, C_k \subseteq C_j$$ and suppose all $C_i$ are closed, connected and non-empty.
Then $\bigcap_{i \in I} C_i$ is also (closed and) connected.
I think this is really the statement that your notes should contain. Or maybe a variation where $I$ is ordered and $C_i$ are decreasing for increasing $i$, which is a bit less general.
(Added after clarification by OP):
We define the quasicomponent of $x$ in any space $X$ as $Q_x = \bigcap \{C : C \text{ clopen in }X, x \in C\}$. This intersection is well-defined as we at least have $X$ itself as such a clopen set.
It's clear that if $C_x$ is the connected component of $x$, and $C$ is any clopen set containing $x$, then $C_x \subset C$, as otherwise some $p \in C_x\setminus C$ would exist and as $x \in C \cap C_x$, $C$ and $X\setminus C$ would disconnected $C_x$ into two non-empty clopen sets. As this holds for all such clopen sets, $C_x \subseteq Q_x$.
The claim now is that for compact and Hausdorff $X$, we have that $Q_x$ is connected and by maximality of the component and that $x \in P_x$ we can then conclude that $Q_x \subseteq C_x$ as well, and so components are equal to quasicomponents.
Now, assume that $Q_x$ were disconnected, so that $Q_x$ is a disjoint union of two relatively clopen sets $C$ and $D$. It's true that one of these contains $x$ (say $C$ does) but we cannot as yet conclude that $C$ is one of the sets in the intersection that forms $Q_x$, because $C$ is merely clopen in $Q_x$, not in $X$ (it is closed in $X$ as $P_x$ is also closed and closed in closed is closed, but not open). But we do have that there is some $O$, open in $X$ so that $O \cap P_x = C$. We're going to use that $C$ and $D$ are disjoint
closed sets in the normal space $X$ (from compact and Hausdorff), and so we have open disjoint subsets $U$ and $V$ of $X$ such that $C \subset U, D \subset V$.
Then by a standard fact on intersection of compact sets (or closed sets in a compact space) we get that there are finitely many clopen subsets $C_1,\ldots C_n$ such that $C_0 := \cap_{i=1}^n C_i \subseteq U \cup V$.
Show that $C_0 \cap U$ is in fact clopen in $X$, and then $x \in C_0 \cap U$ shows that $Q_x \subseteq C_0 \cap U \subseteq U$. This then implies $D$ is empty and $Q_x$ is connected.
