The number of subsets of $\{1, 2, \dots, 2n\}$ that have $k$ as smallest element is $2^{2n-k}$ for $1 \le k \le 2n$ since each element bigger than $k$ is either contained in the subset or not.
The number $O$ of subsets with an odd smallest element is therefore equal to
$$O = 2^{2n-1} + 2^{2n-3} + \dots + 2^3 + 2^1 = 2 \cdot (4^{n-1} + 4^{n-2} + \dots + 4^1 + 4^0).$$
The number $E$ of subsets with an even smallest element is equal to
$$E = 2^{2n-2} + 2^{2n-4} + \dots + 2^2 + 2^0 = 4^{n-1} + 4^{n-2} + \dots + 4^1 + 4^0.$$
This implies $O = 2E$ and consequently the desired proportion is $2/3$.
(Birgit Vera Schmidt) $\square$