Use induction to prove that a finite set with n elements has 2^n subsets.
See here:
http://www.cs.uiuc.edu/class/fa08/cs...es/lect_17.pdf
Google your problem. It's found all over the place.
The statement holds for $\displaystyle n=1$: $\displaystyle \emptyset \subseteq \left\{ 1 \right\}\;\& \;\left\{ 1 \right\} \subseteq \left\{ 1 \right\}$.
Now suppose that $\displaystyle S = \left\{ {1,2, \cdots ,k} \right\}$ and $\displaystyle S$ has $\displaystyle 2^k$ subsets.
Consider $\displaystyle S^+=S \cup \left\{ {k + 1} \right\}$. Each subset of $\displaystyle S$ is also a subset of $\displaystyle S^+$ this is $\displaystyle 2^k$ subsets. Unite each subset of $\displaystyle S$ with $\displaystyle \left\{k+1\right\}$ giving us $\displaystyle 2^k$ subsets more subsets. So we have $\displaystyle 2^k + 2^k = 2\left( {2^k } \right) = 2^{k + 1} $ subsets of $\displaystyle S^+$. That completes the induction.