Does removing finitely many points from an open set yield an open set?
The argument makes sense in general topological spaces where singletons are closed ($T_1$ spaces I think).
If some singleton $\{x\}$ is not closed in the space $X$, then $X$ is open in $X$ while $X \backslash \{x\}$ is not.
If you allow arbitrary topologies ("any general space"), then the answer is no. For any set $M$ with at least two elements you can define a topology $\mathcal T=\{\emptyset, M\}$ (i.e. only the empty set and $M$ are open sets in $M$).
Then, by removing one point from $M$, you get a set which is not open.
Edit: Let me add the following:
Proposition. Let $(M,\mathcal T)$ be a topological space. Then the following two statements are equivalent:
- For every $x\in M$, $\{x\}$ is closed (i.e. $M\setminus\{x\}\in\mathcal T$),
- For every open set $O\in \mathcal T$ and point $x\in M$, we have that $O\setminus\{x\}$ is also open.
Proof. 1.$\implies$2.: Suppose that 1. is true and let $O\in \mathcal T$, $x\in M$. Then $M\setminus\{x\}$ is open and hence $$O\setminus\{x\}=O\cap\big(M\setminus\{x\}\big)\in\mathcal T.$$
2.$\implies$ 1.: Suppose that 2. is true and let $x\in M$. Then, since $M$ is open, $M\setminus\{x\}$ is open too, so $\{x\}$ is closed. $\square$
The reason this works in $\Bbb R^n$ is that a set consisting of a single point is always closed. Then you can understand $$G\setminus\{ x_1, x_2, \ldots, x_k \}$$ as $$G\cap (\Bbb R^n \setminus \{x_1\}) \cap (\Bbb R^n \setminus \{x_2\})\cap\ldots \cap (\Bbb R^n \setminus \{x_k\}).$$ This is a finite intersection of open sets, so is open.
In general spaces, a singleton set $\{x_1\}$ may not always be closed. A space in which every singleton is closed is called a $T_1$ space. All metric spaces are $T_1$, but it is easy to construct quite simple spaces that are not $T_1$. For example, let $S=\{a,b,c\}$ and let the open sets be $S, \{a,b\}, \{a,c\}, \{a\}, \varnothing$. Then $\{c\}$ is closed but $\{a\}$ is not, and indeed if you remove the single point $a$ from the open set $\{a,b\}$ you get $\{b\}$, which is not an open set.