I found a Abstract algebra book in goodwill

I can not get how do prove something. It seems almost impossible in some cases.

This is a question in it:

Prove : $\displaystyle (A\cap B)'= A' \cup B'$

Where $\displaystyle n'=U\setminus n$

and U is the universal set.

Here is my attempted proof:

$\displaystyle (A \cap B)' \Rightarrow$

$\displaystyle x\in (U\setminus (A\cap B) \Rightarrow$

$\displaystyle (x\in U) \wedge (x\notin (A\cap B)) \Rightarrow$

$\displaystyle (x\in U) \wedge (x\notin A \wedge x\notin B) \Rightarrow$

$\displaystyle (x\in U \wedge x\notin A) \vee (x\in U \wedge x\notin B) \Rightarrow$

$\displaystyle (x\in (U \setminus A) \vee x\in(U\setminus B) \Rightarrow$

$\displaystyle x\in (U\setminus A \cup U\setminus B) \Rightarrow$

$\displaystyle x\in A' \cup B'$

$\displaystyle \therefore\,\, A'\cup B' \subseteq (A\cap B)'$

And the reverse is true. Therefore they must be equal.

Is this correct?