Let us introduce the following definition:
Let $X$ be a set. A mapping $f: \mathcal{P}(X) \rightarrow \mathcal{P}(X)$ is said to be a closure operator if for all subsets $S, T \subseteq X$, the following holds:
$S \subseteq f(S), \quad$ ($f$ is extensive)
$S \subseteq T\Rightarrow f(S) \subseteq f(T), \quad$ ($f$ is isotone)
$f(f(S)) = f(S),\quad$ ($f$ is idempotent)
Now, let $C \subseteq X$ be a subset such that $f(C)=C$ and let $p \in C$ be a point such that $p \notin f(C \setminus \left \{p \right \})$. I want to prove that $f(C \setminus \left \{p \right \})=C \setminus \left \{p \right \}$.
I tried proving it by (way of) contradiction:
Assume that $f(C \setminus \left \{p \right \}) \neq C \setminus \left \{p \right \}$. By the extensivity of $f$, it holds that $f(C \setminus \left \{p \right \}) \supset C \setminus \left \{p \right \}$, where $\supset$ denotes a proper superset. Furthermore, by the isotonity of $f$, it holds that $f(C \setminus \left \{p \right \}) \subseteq f(C)=C$.
Here is where I am not sure if I can proceed. From the relation $C \setminus \left \{p \right \} \subset f(C \setminus \left \{p \right \}) \subseteq C$, it follows that $f(C \setminus \left \{p \right \})=C$ which is a contradiction since we required $p \notin f(C \setminus \left \{p \right \})$, but $p \in C$. Therefore, it holds that $f(C \setminus \left \{p \right \})=C \setminus \left \{p \right \}$.
Is my proof correct?
Another way of putting this could be the following:
$$C\setminus{\{p\}} \subseteq f\big(C\setminus{\{p\}}\big) \subseteq f(C) = C\quad\text{ implies}$$
$$\big(C\setminus{\{p\}}\big)\setminus\{p\}\subseteq f\big(C\setminus{\{p\}}\big)\setminus\{p\}\subseteq C\setminus\{p\}\;.$$
Since $\;p\not\in f\big(C\setminus\{p\}\big)\,,\,$ it follows that
$$C\setminus{\{p\}} \subseteq f\big(C\setminus{\{p\}}\big) \subseteq C\setminus\{p\}\quad$$ hence
$$f\big(C\setminus{\{p\}}\big)=C\setminus\{p\}\,.$$
Actually the hypothesis $\,p\in C\,$ is not necessary, in fact in this proof we have not used it.