Question: Given the topological space $\Omega$, let $\chi_A: X \rightarrow \mathbb{R}$ be the characteristic function of $A\subseteq \Omega$ (hence $\chi_A(x)=1$ iff $x \in A$ and $0$ otherwise). Also suppose $\mathbb{R}$ is under the standard topology.
Given $x_0 \in X$ prove there is an open set $U$ containing $x_0$ such that $\chi_A\restriction_U$ is continuous if and only if $x_0 \notin cl(A) \cap cl(X \backslash A)$
Attempt: If $U \subset A$ and $x_0$ in $U$ then clearly $x_0 \notin cl(X \backslash A)$. Similarly if $U \subset X \backslash A$ then $x_0 \notin cl(A)$. Thus the interesting case is when $U \cap A\neq \emptyset$ and $U \cap X \backslash A \neq \emptyset$ but I can't figure out how to get started...
If every neighbourhood $U$ of $x_0$ is such that $U \cap A \neq \emptyset$ and $U \cap (X\setminus A) \neq \emptyset$ this exactly means that $x_0 \in \operatorname{cl}(A) \cap \operatorname{cl}(X\setminus A)= \partial A$ and then $\chi_A\restriction_U$ is not continuous (at $x_0$). In the other case, as you also said, either $x_0 \in \operatorname{int}(A)$ or $x \in \operatorname{int}(X\setminus A)$ and we can take these interiors as an open set $U$ on which $\chi_A$ is constant and thus certainly continuous.