Let $\{B_i:i=1,\cdots,n\}$ be a partition of $\Omega$ so that $B_i\cap B_j=\emptyset$, and $\bigcup_{i=1}^n B_i =\Omega$. Define $\mathscr{G}=\sigma\{B_i:i=1,\cdots,n\}$. Then $$\{X\in \mathscr{G}:EX^2<\infty\}=\{\sum_{i=1}^n c_iI_{B_i}\}.$$
I know that $\mathscr{G}=\{\bigcup_{i\in J}B_i:J\subseteq\{1,\cdots,n\}\}$. Then how to go on the proof?
I preassume that $X\in\mathscr{G}$ must state here that $X$ is measurable wrt $\sigma$-algebra $\mathscr{G}$.
(It is not a notation that I am familiar with.)
$X=\sum_{i=1}^nX\mathbf1_{B_i}$ and if $X$ is measurable then for every index $i$ also $X\mathbf1_{B_i}$ is measurable.
Now suppose that $X(\omega_1)=c_1\neq c_2=X(\omega_2)$ for $\omega_1,\omega_2\in B_i$.
Since $c_1\neq c_2$ it is not possible that both equalize $0$ and WLOG we may assume that $c_1\neq 0$.
Then $(X\mathbf1_{B_i})^{-1}(\{c_1\})$ must be a non-empty measurable proper subset of $B_i$.
This is not possible in this context, so we conclude that $X$ is constant on $B_i$.
If $X$ takes value $c_i$ on $B_i$ then we get: $$X=\sum_{i=1}^nc_i\mathbf1_{B_i}$$