Proving the total number of subsets of S is equal to $2^n$

You can prove it by induction.

First off start by noticing that $$ \displaystyle \sum_{r=1}^n \frac{n!}{r!(n-r)!} = 2^n - 1\iff \displaystyle \sum_{r=0}^n \overbrace{\frac{n!}{r!(n-r)!}}^{\displaystyle \binom{n}{r}} = 2^n$$

Due to the equivalence above, it suffices to prove that $\displaystyle \sum_{r=0}^n \binom{n}{r} = 2^n$. (The only reason I chose to prove this one instead is because I feel it is more aesthetically pleasing).

Base case: $n=0$

$\displaystyle \sum_{r=0}^0 \binom{0}{r} = \displaystyle \sum_{r=0}^0 \frac{0!}{r!(0-r)!} = \frac{0!}{0!0!}=1=2^0$

Induction step: Let $n\in \Bbb N$ and suppose $\displaystyle \sum_{r=0}^n \binom{n}{r} = 2^n$. Multiply by $2$ to get: $$\displaystyle 2\sum_{r=0}^n \binom{n}{r} = 2^{n+1}.$$

Also that is left is to prove that $\displaystyle 2\sum_{r=0}^n \binom{n}{r}=\sum_{r=0}^{n +1}\binom{n+1}{r}$. The green equality below is a consequence of Pascal's rule:

$$\begin{align} 2\sum_{r=0}^n \binom{n}{r}&=\sum_{r=0}^n \binom{n}{r}+\sum_{r=0}^n \binom{n}{r}\\ &=\binom{n}{0}+\sum_{\color{red}{r=1}}^n \binom{n}{r}+\sum_{r=0}^{\color{blue}{n-1}} \binom{n}{r}+\binom{n}{n}\\ &=1+\sum_{\color{red}{r=0}}^\color{red}{n-1} \binom{n}{\color{red}{r+1}}+\sum_{r=0}^{{n-1}} \binom{n}{r}+1\\ &=1+\sum _{r=0}^{n-1}\left[\binom{n}{r+1}+\binom{n}{r}\right]+1\\ &\color{green}=1+\sum_{r=0}^{n-1} \binom{n+1}{r+1} + 1\\ &=\binom{n+1}{0}+\sum _{\color{red}{r=1}}^\color{red}n\binom{n+1}{\color{red}r}+\binom{n+1}{n+1}\\ &=\binom{n+1}{0}+\sum _{r=1}^{n+1}\binom{n+1}{r}\\ &=\sum _{r=0}^{n+1}\binom{n+1}{r} \end{align}$$

Note that starting on the third line I use $n-1$. This is only correct if $n\ge 1$, but if $n=0$, it's just the base case.

I just noticed that your base case is for $n=1$, that's fine. Mine is a bit more general.

Hint: To form a subset, for each element of $S$ we have to decide whether to include or to exclude that element. If $|S| = n$, $n$ decisions must be made.

This can be proved by simple approach as : Suppose S is $ S = \lbrace a_1,a_2,a_3,.............a_n \rbrace $ For subset a element can be selected in 2 way,either it is selected or not selected. So total subset is :

$ \lbrace 2.2.2.2............2\rbrace $ \\ n times multiplication of 2

$ = 2^n $