I never encountered a confirming answer on the question in the title so have great distrust in my own reasoning below.
That made me decide to ask it as a question here.
Let $X$ be a topological space with $x\in X$.
Let $\mathsf{CO}\left(X\right)$ denote the collection of clopen subsets of $X$.
Equipped with $\cap,\cup$ and complement $\mathsf{CO}\left(X\right)$ can be recognized as a Boolean algebra.
Let $u_{x}=\left\{ C\in\mathsf{CO}\left(X\right)\mid x\in C\right\} $.
Then $u_{x}$ can be recognized as an ultrafilter of Boolean algebra $\mathsf{CO}\left(X\right)$.
We have $x\in\bigcap u_{x}$ so that $\bigcap u_{x}\neq\varnothing$.
An ultrafilter that is not free must be a principal filter generated by an atom, so some atom $A\in\mathsf{CO}\left(X\right)$ must exist with $u_{x}=\left\{ C\in\mathsf{CO}\left(X\right)\mid A\subseteq C\right\} $.
Then $\bigcap u_{x}=A\in\mathsf{CO}\left(X\right)$ showing that $\bigcap u_{x}$ is an open set.
But $\bigcap u_{x}$ can also be recognized as the quasi-component represented by element $x$.
Proved is now that quasi-components are open sets.
I must have made a mistake somewhere, I think, and will be thankful if you could point out what I did wrong.
You have defined an ultrafilter in a Boolean algebra, agreed. But $\bigcap u_x$ (or really $\bigwedge u_x$) is not defined in that BA necessarily. Having it non-empty as sets means little. Assuming the intersection is clopen is saying that $\mathrm{CO}(X)$ is complete; you only have $\emptyset=0$ as a lower bound for $u_x$ in $\mathrm{CO}(X)$, so if the inf existed it could well be $\emptyset$ too. Claiming $u_x$ is not free in the BA is already positing the existence of a clopen $A\neq \emptyset$ as a lowerbound, begging the question.
Also, $\mathrm{CO}(X)$ could very well be atomless, so the "generated by an atom" statement makes no sense.