Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space

Let $K \subseteq X$ be clopen with $x \in K$. Then $K \cap C(x)$ is both open and closed in the subspace $C(x)$ by [A3], and it contains $x$, so it is nonempty.

Every clopen $K \ni y$ also contains $x$ whenever $y \in Q(x)$: otherwise $X \setminus K$ is a clopen set containing $x$ by [A4], so $Q(x) \subseteq X \setminus K$ by [A1], contradicting $y \in Q(x) \cap K$.

$Q(x)$ is closed, being by [A1] the intersection of a nonempty family of clopen, hence closed, sets, and such an intersection is closed by [A4]; this is claim 2. And $x \in Q(x)$, every member of that family containing $x$.

Since $C(x)$ is connected by [A1], its only clopen subsets are $\varnothing$ and $C(x)$ by [A2]; so step 1.1 forces $K \cap C(x) = C(x)$, that is $C(x) \subseteq K$.

Let $y \in Q(x)$. Every clopen $K$ with $x \in K$ contains $y$, since $Q(x) \subseteq K$ by [A1], so $Q(y) \subseteq K$; hence $Q(y) \subseteq Q(x)$. Conversely every clopen $K$ with $y \in K$ contains $x$ by step 1.2, so $Q(x) \subseteq K$ and therefore $Q(x) \subseteq Q(y)$. Thus $Q(y) = Q(x)$.

As $K$ was an arbitrary clopen set containing $x$, it follows that $C(x)$ is contained in the intersection of all of them, that is $C(x) \subseteq Q(x)$; this is claim 1.

So for $y \in Q(x)$ one has $C(y) \subseteq Q(y) = Q(x)$ by step 3.1 and step 2.2; and each such $y$ lies in $C(y)$ by [A1], so $Q(x) = \bigcup \{\, C(y) : y \in Q(x) \,\}$. This is claim 3.

For claim 4: each $Q(x)$ is nonempty by step 1.3; if $z \in Q(x) \cap Q(y)$ then $Q(z) = Q(x)$ and $Q(z) = Q(y)$ by step 2.2, so $Q(x) = Q(y)$, and hence two quasicomponents are equal or disjoint; and $x \in Q(x)$ by step 1.3, so they cover $X$.