Proof of inclusion-exclusion

Indicators. For a set $A$, $1_A(x)=1$ if $x\in A$, $0$ otherwise. Then $1_{\bigcup A_i}(x)=1$ iff some $1_{A_i}(x)=1$, i.e. $1-\prod_i(1-1_{A_i}(x))$.

Expand. $\prod_i(1-1_{A_i})=\sum_I(-1)^{|I|}1_{\bigcap_{i\in I}A_i}$ (with $I=\emptyset$ giving $1$).

Thus. $1_{\bigcup A_i}=\sum_{\emptyset\neq I}(-1)^{|I|+1}1_{\bigcap A_i}$.

Sum over $x$. Summing both sides: $|\bigcup A_i|=\sum_{\emptyset\neq I}(-1)^{|I|+1}|\bigcap_{i\in I}A_i|$.

Conclusion. The formula holds.