In a compact Hausdorff space every quasicomponent is connected, so quasicomponents and components coincide

Write $Q := Q(x)$ and suppose $Q$ carries a separation: disjoint nonempty sets $A, B$, open in the subspace $Q$, with $A \cup B = Q$, and $x \in A$ after renaming, since $x \in Q$ by [L1]. By [L4] each of $A$ and $B$ is also closed in $Q$; $Q$ is closed in $X$ by [L2], so $A$ and $B$ are closed in $X$ by [L5] and compact subsets of $X$ by [L6].

By [L7] there are disjoint open $U \supseteq A$ and $V \supseteq B$ in $X$, and then $Q = A \cup B \subseteq U \cup V$.

Let $\mathcal{K}$ be the family of clopen subsets of $X$ containing $x$ and put $\mathcal{F} := \{\, K \setminus (U \cup V) : K \in \mathcal{K} \,\}$, a family of closed subsets of $X$ by [L9]. Its intersection is $Q \setminus (U \cup V)$ by [L1], which is empty by step 2.1.

By [L8] the family $\mathcal{F}$ therefore fails the finite intersection property, so some finite list in it has empty intersection; the empty list is not such a list, its intersection being $X$, which contains the nonempty $A$. So there are $n \in \mathbb{N}$ and $K_0, \dots, K_n \in \mathcal{K}$ with $(K_0 \cap \dots \cap K_n) \setminus (U \cup V) = \varnothing$, and $K := K_0 \cap \dots \cap K_n$ is a clopen set containing $x$ with $K \subseteq U \cup V$.

$K \cap U$ is clopen: it is open as an intersection of two open sets, and it equals $K \setminus V$, since $K \subseteq U \cup V$ and $U \cap V = \varnothing$, so it is the intersection of the closed $K$ with the closed complement of $V$. It contains $x$, because $x \in A \subseteq U$ and $x \in K$.

So $K \cap U$ belongs to $\mathcal{K}$ and [L1] gives $Q \subseteq K \cap U \subseteq U$; but $B$ is a nonempty subset of $Q$ contained in $V$, so it lies in $U \cap V = \varnothing$. This is impossible, so $Q$ admits no separation and is connected by [L4].

Hence $Q(x)$ is a connected subset of $X$ containing $x$, so $Q(x) \subseteq C(x)$ by [L3], while $C(x) \subseteq Q(x)$ by [L2]; the two sets are equal, and since every component and every quasicomponent is of the form $C(y)$ and $Q(y)$ for a point $y$, the two families coincide.