1. Lemma: We start by proving the lemma:
$$\binom{2n}{n} = 2^n \cdot \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \ge 2 \cdot 3^{n-1}$$
Proof: The proof of this lemma can be found here. For the second part, note that $2k-1 > \frac{3}{2}k$ for $k > 1$. This inequality can be used to show that the product of odd numbers up to $2n-1$ is greater than a certain bound.

2. Problem Setup: Let $x_i$ (for $1 \le i \le n$) be the number of two-element sets among the $2n-1$ sets that $i$ belongs to. Note that $x_i \le n-1$.

3. Random Selection and Probability Calculation: Pick $n$ sets randomly from the $2n-1$ sets. The probability that a particular element $i$ is in at least one of the chosen sets is:
$$\mathbb{P}(X_i) = \frac{\sum_{j=1}^n \binom{x_i}{j}}{\binom{2n-1}{n}} \le \frac{2^{n-1} - 1}{\binom{2n-1}{n}}$$
Here, $X_i = 1$ if $i$ belongs to at least one of the chosen sets, and $X_i = 0$ otherwise.

4. Expected Number of Elements in the Union: The expected number of elements in the union of $n$ sets is:
$$\mathbb{E}[X] \le \frac{n(2^{n-1} - 1)}{\binom{2n-1}{n}}$$

5. Inequality and Bound Calculation: It suffices to prove that:
$$2^{n-1} - 1 < \binom{2n-1}{n} \cdot k$$
for the best $k$ possible. This is equivalent to:
$$2^n - 2 < \binom{2n}{n} \cdot k$$
Using the lemma, we have:
$$\binom{2n}{n} \ge 2 \cdot 3^{n-1}$$
Therefore, we need:
$$2^n - 2 < 2 \cdot 3^{n-1} \cdot k$$
Solving for $k$, we find that $k = \frac{2}{3}$ is sufficient. Thus:
$$\mathbb{E}[X] \le \frac{2}{3}n$$
This means there exists $n$ subsets such that their union has at most $\frac{2}{3}n$ elements.

6. Conclusion: Therefore, we have shown that one can choose $n$ out of the $2n-1$ two-element subsets such that their union contains no more than $\frac{2}{3}n + 1$ elements.

$\blacksquare$