Consider the probability space as bellow so that $A_1=A_2$:
What is wrong in the following proof:
\begin{equation} P(B)=\frac{1}{2}(P(B|A_1)+P(B|A_2))\\ 2P(B)-2P(B|A_1)=P(B|A_2)-P(B|A_1)\\ 2P(B)-2P(B|A_2)=P(B|A_1)-P(B|A_2) \end{equation} Since $P(B)\geq P(B|A_1)$ and $P(B)\geq P(B|A_2)$: \begin{equation} P(B|A_2)\geq P(B|A_1)\\ P(B|A_1)\geq P(B|A_2) \end{equation} Hence $$P(B|A_1)= P(B|A_2)$$
In your first line, you probably mean that $P(A_1)=P(A_2)=1/2$, since you use this fact later on. This clearly does not mean that $A_1=A_2$, as your image illustrates.
Your mistake is writing that $P(B)\ge P(B\,|\,A_1)$. This does not hold for any sets $A_1,B\subset\Omega$. What does hold is $P(B)\ge P(B\cap A_1)$ because $B\cap A_1\subset B$.
As a counter example, take $A_1=B$, in which case you have $P(B\,|\, B)=1\ge P(B)$ with the inequality being strict for some $B$.