Consider a probability space $(\Omega,\mathcal{F},P)$ and a bounded random variable $X$. Let $(A_n)_{n\ge 1}$ be a countable partition of $\Omega$ and define $\mathcal{A}:=\sigma(\{A_n:n\ge 1\})$. Then it holds:
$$E[X|\mathcal{A}]=\sum_{n\ge 1, P(A_n)>0}E[X|A_n]\cdot\mathbb{1}_{A_n}$$
One can prove it by using the uniqueness of $E[X|\mathcal{A}]$, i.e. I should be done, if I can show that the expression on the right is $\mathcal{A}$-measurable, and that for any $B\in\mathcal{A}$, it holds
$$E\Big[E[X|\mathcal{A}]\cdot 1_B\Big]=E\Big[\Big(\sum_{n\ge 1, P(A_n)>0}E[X|A_n]\cdot\mathbb{1}_{A_n}\Big)\cdot 1_B\Big]$$
EDIT:
First property: For any $n\ge 1$: $E[X|A_n]$ and $1_{A_n}$ are $\sigma(1_{A_n})$-measurable, therefore $\mathcal{A}$-measurable. Now using the fact that products and sums of measurable functions are measurable, we have that
$$\sum_{n\ge 1, P(A_n)>0}E[X|A_n]\cdot\mathbb{1}_{A_n}$$
is $\mathcal{A}$-measurable.
Second property:
Now let $A_n\in\mathcal{A}$ with $P(A_n)=0$. Then we know that
$$E[X\cdot1_{A_n}]=\sum_{\omega\in A_n}X(\omega)\cdot \underbrace{P(\{\omega\})}_{=0}=0.$$
Therefore we have for any $B\in\mathcal{A}$, i.e. $B=\cup_{m\in M} A_m$ with a countable subset $M\subset\mathbb{N}$
\begin{align}
E\Big[E[X|\mathcal{A}] 1_B \Big]
&=E[X 1_B]\\
&=E[X\sum_{m\in M}1_{A_m}]\\
&=E[\sum_{m\in M}X 1_{A_m}]\\
&=\sum_{m\in M}E[X 1_{A_m}],\text{ since $X$ is bounded we can use Fubini here}\\
&=\sum_{m\in M,P(A_m)>0}E[X 1_{A_m}],\text{ with the statement above}\\
&=\sum_{m\in M,P(A_m)>0}E\Big[E[X|A_m] 1_{A_m}\Big]\\
&=\sum_{m\in M,P(A_m)>0}E\Big[E[X|A_m] 1_{A_m}1_B\Big],\text{ since }1_B\cdot 1_{A_m}=1_{A_m}.\\
&=\sum_{n\ge 1,P(A_n)>0}E\Big[E[X|A_n] 1_{A_n}1_B\Big],\text{ since }1_B\cdot 1_{A_n}=0\text{ for }n\notin M.\\
&=E\Big[\Big(\sum_{n\ge 1,P(A_n)>0}E[X|A_n] 1_{A_n}\Big)1_B\Big],\text{ again Fubini}\\
\end{align}
For the Fubini statement see here
Best Answer
Assume for simplicity that $\mathsf{P}(A_i)>0$ for all $i$. Every element of $\mathcal{A}$ is an at most countable union of sets in $(A_n)$. Thus, for $B=\bigcup_{i}A_i$, $$ \mathsf{E}X1_B=\sum_{i}\mathsf{E}[X1_{A_i}]=\sum_{i}\mathsf{E}[\mathsf{E}[X\mid A_i]1_{A_i}]=\mathsf{E}\left[\sum_i\mathsf{E}[X\mid A_i]1_{A_i}1_B\right] $$ because $(A_n)$ are disjoint and $X$ is bounded (although, it suffices to assume that $\mathsf{E}|X|<\infty$).