Proving 2 properties of sigma-algebra Let $\mathcal A \subset 2^{Ω}$, $Ω \neq \emptyset$. 
Also, $\mathcal A''=\cap_{\mathcal A \subset \mathcal A'} \mathcal A'$, where $\mathcal A' \subset 2^{Ω}$ is sigma-algebra.
Show that:
(a) If $A \in \mathcal A'',$ then $A^{c} \in \mathcal A''.$
(b) If $A_1, A_2, ..., A_n, ... \in \mathcal A'', $ then $\cup_{n=1}^{\infty} A_n \in \mathcal A''.$
I started from:
From the definition: the smallest $σ–$algebra containing $\mathcal A$ is $σ(\mathcal A)$=$\mathcal A''=\cap_{\mathcal A \subset \mathcal A'} \mathcal A'.$
 A: You "started from definition" but the thing is that implicitly you are asked to prove that "a smallest $\sigma$-algebra containing $\mathcal A$" exists and can be identified with the collection $\mathcal A''$ that is mentioned in your question. In that stage you cannot start with it.

Let $\mathfrak A:=\{\mathcal A'\mid \mathcal A'\text{ is a }\sigma\text{-algebra that satisfies }\mathcal A\subseteq\mathcal A'\subseteq\wp(\Omega)\}$.
Then $\mathcal A''=\cap\mathfrak A$.
a) 
$A\in\mathcal A''$ is the same statement as: for every $\sigma$-algebra $\mathcal A'$ with $\mathcal A\subseteq\mathcal A'\subseteq\wp(\Omega)$ we have $A\in\mathcal A'$.
From this it follows directly that: for every $\sigma$-algebra $\mathcal A'$ with $\mathcal A\subseteq\mathcal A'\subseteq\wp(\Omega)$ we have $A^{\complement}\in\mathcal A'$.
Or equivalently: $A^{\complement}\in\mathcal A''$.
b)
$A_1,A_2,\dots\in\mathcal A''$ is the same statement as: for every $\sigma$-algebra $\mathcal A'$ with $\mathcal A\subseteq\mathcal A'\subseteq\wp(\Omega)$ we have $A_1,A_2,\dots\in\mathcal A'$.
From this it follows directly that: for every $\sigma$-algebra $\mathcal A'$ with $\mathcal A\subseteq\mathcal A'\subseteq\wp(\Omega)$ we have $\bigcup_{n=1}^{\infty}A_n\in\mathcal A'$.
Or equivalently $\bigcup_{n=1}^{\infty}A_n\in\mathcal A''$.
This together proves that $\mathcal A''$ is closed under complements and countable unions. Further the collection is not empty so that we are allowed to conclude that $\mathcal A''$ is a $\sigma$-algebra. This with the special property that it contains $\mathcal A$ and that it is a subcollection of any $\sigma$-algebra $\mathcal A'$ that contains $\mathcal A$ (so is the smallest $\sigma$-algebra that contains $\mathcal A$).
Now we are justified to define $\sigma(\mathcal A)$ as $\mathcal A''$ being the smallest $\sigma$-algebra that contains $\mathcal A$.
So we end where you started.
