A question in probability theory

We prove this by contradiction.

Suppose there are no sets $A \in \mathcal{A}$ with $P(A)=a.$ Define
\[\mathcal{B}=\{\hat{B} \in \mathcal{A}: P(\hat{B})\leq a\}.\]
Note that $\mathcal{B}$ is inductive under inclusion, i.e.
\[\hat{B_1}\leq \hat{B_2}   \text{iff}    \hat{B_1} \subset\hat{B_2}.\]
Thus by Zorn's lemma (every partially ordered set for which every totally ordered subset has an upper bound contains at least one maximal element) $\mathcal{B}$ has a maximal element $\hat{B}_0\in \mathcal{B}$ with $P(\hat{B}_0 )\leq a$. Since we have assumed that there is no set with probability $a$, we must have 
\[P(\hat{B}_0 )<a.\]

Next define 
\[\mathcal{C}=\{\hat{C}\in \mathcal{A}:  \hat{C}\neq \emptyset,  P(\hat{C})\geq 0  \text{and}  \hat{C}\cap \hat{B}_0=\emptyset\}.\]
Again $\mathcal{C}$ is inductive under inclusion, i.e.
\[\hat{C_1}\leq \hat{C_2}    \text{iff}      \hat{C_2} \subset \hat{C_1}.\]
Thus by Zorn's lemma it has a maximal element $\hat{C}_0 \in \mathcal{C}$ with 
\[P(\hat{C}_0)\geq 0.\]

Now we consider two cases: 

Case I: $P(\hat{C}_0)=0$. Since $\hat{C}_0 \neq \emptyset$, and $\hat{C}_0 \cap \hat{B}_0=\emptyset$, we have
\[\hat{B}_0 \subset B^{*}:=\hat{B}_0\cup \hat{C}_0 \]
and 
\[P(B^*)=P(\hat{B}_0)+P(\hat{C}_0)=P(\hat{B}_0)\leq a.\]
Thus $B^* \in \mathcal{B}$ and $\hat{B}_0 \subset B^*$. This contradicts the assumption that $\hat{B}_0$ is maximal.

Case II: $P(\hat{C}_0)>0$. Since $\hat{C}_0$ is maximal, $\hat{C}_0$ must be an atom, otherwise we can remove an element from $\hat{C}_0$ and get another set $C^*$ with 
\[C^* \geq \hat{C}_0,\]
which contradicts the assumption that $\hat{C}_0$ is maximal.

Both cases I and II lead to a contradiction. Therefore, there should exists a set $A\in \mathcal {A}$ with $P(A)=a.$ 

Note: To use the Zorn's lemma a set must be partially ordered. In the context above we have
\[\hat{B_1}\subset \hat{B}_2    \text{and}    \hat{B_2}\subset \hat{B}_3   \Rightarrow  \hat{B_1}\subset \hat{B}_3,\]
which means that inclusion is partially ordered, i.e.
\[\hat{B_1}\leq \hat{B}_2   \text{and}   \hat{B_2}\leq \hat{B}_3   \Rightarrow   \hat{B_1}\leq \hat{B}_3.\]
This is what it means to say "set to be inductive under inclusion" in this context. You can also see that $\mathcal{C}$ is a partial order. You just need to replace $\subset$ with $\supset$. I suggest you to review Zorn's lemma: 

https://en.wikipedia.org/wiki/Zorn%27s_lemma