Connectedness of compact metric space Let $X$ be a compact metric space satisfying the following condition: for any given positive number $\delta>0$, only finitely many components of $X$ have diameter larger than $\delta$.
For a given component $P$ of $X$, let us remove a connected compact subset $C$ of $P$ such that $P\setminus C$ is disconnected. Consider a disjoint partition $P\setminus C=A\cup B$, with $A$ and $B$ at a positive distance. Suppose that $x\in A$ and $y\in B$.
Are then $x$ and $y$ in different quasi-components of $X\setminus C$? or can we find a separation of $X\setminus C=M\cup N$ such that $x\in M$ and $y\in N$?
 A: (This answers the previous version of the question, for components but not quasicomponents).
I think, it us true in every topological space: if $X$ is a topological space with connected components $P$ and $Q_j$: $X=P\sqcup (\sqcup_j Q_j)$, and you remove  a set $C$ from a connected component $P$ so that it becomes disconnected: $P\setminus C=\sqcup A_i$, where $A_i$ are the connected components of $P\setminus C$, then the connected components of $X\setminus C$ are $A_i$'s and  $Q_j$'s. The reason is that the connected components are the inclusion-maximal connected subsets, and each connected subset of $X\setminus C$ lies inside a single connected component of $X$.
A: Your compactum $X$ is finitely Suslinian (by defnition). As such, every connected subset of $X$ is locally connected (see this paper), and this implies that the components of $X\setminus C$ are the same as the quasi-components of $X\setminus C$.  See this paper, Theorem 2.1. @Fedor Petrov already argued that $x$ and $y$ are in different components of $X\setminus C$, therefore they are also in different quasi-components.
Note that $C$ does not have to be connected or compact.  It can be any subset of $P$ such that $P\setminus C$ is disconnected.
