A bijective proof

Bijection. Map $S\subseteq\{1,\ldots,n\}$ with $|S|=k$ to $\bar S$, its complement, of size $n-k$.

Injective/surjective. The complement map is its own inverse, hence bijective.

Conclusion. $|\{$k-subsets$\}|=|\{$(n-k)-subsets$\}|$, i.e. $\binom{n}{k}=\binom{n}{n-k}$.