Let $(X,\tau)$ be a topological space, and let $Q$ be a quasi-component of $X$. Let $S$ be a subset of $X\setminus Q$. Then is $Q$ necessarily a quasi-component of $X$ in the topology generated by $\tau\cup\{S\}$?



Slightly finer topology vs a quasi-component
Let $X$ be the subspace of the plane given by $X = \{ (\frac{1}{n},y) : n = 1, 2, \cdots,\ 0 \leq y \leq 1 \} \cup \{(0,0),(0,1)\}$, and let $S = \{ \frac{1}{n} : n = 1, 2, \cdots\} \times \{\frac{1}{2}\}$. Then the quasi-component of $(0,0)$ in $X$ is $\{(0,0),(0,1)\}$ but in the topology generated by $X$ and $S$ the quasi-component of $(0,0)$ is $\{(0,0)\}$. 


In fact, if $X$ is a normal space and $Q$ is a disconnected quasi-component, then there is a subset $S$ of $X \setminus Q$ such that $Q$ is not a quasi-component of the topology generated by adding $S$ to the topology of $X$. For this let $(H,K)$ be a disconnection of $Q$. Then $H$ and $K$ are disjoint closed subsets of $X$. Let $U$ and $V$ be disjoint open subsets of $X$ such that $H \subseteq U$ and $K \subseteq V$. Let $S = X \setminus (U \cup V)$.


