We first prove that $|X| > 2n - 2$. In fact, if $|X| = 2n - 2$, let $X = \{1, 2, \dots, 2n - 2\}$, $B_1 = \{1, 2\}$, $B_2 = \{3, 4\}$, $\dots$, $B_{n-1} = \{2n - 3, 2n - 2\}$. Since $|Y| = n$, there exist two elements in $Y$ that belong to the same $B_i$, then $|Y \cap B_i| > 1$, a contradiction.

Let $|X| = 2n - 1$.
Let $B = \bigcup_{i=1}^n B_i$, then $|B| = 2n - 1 - z$, where $z$ is the number of subset $X \setminus B$. Suppose the elements of $X \setminus B$ are $a_1, a_2, \dots, a_z$.

If $z \ge n - 1$, take $Y = \{a_1, \dots, a_{n-1}, d\}$, and $d \in B$, as desired.

If $z < n - 1$, suppose there are $t$ elements that occur once in $B_1, B_2, \dots, B_n$. Since $\sum_{i=1}^n |B_i| = 2n$, then
$$t + 2(2n - 1 - z - t) \le 2n,$$
it follows that $t \ge 2n - 2 - 2z$. So the elements that occur twice or more than twice in $B_1, B_2, \dots, B_n$ occur repeatedly by $2n - (2n - 2 - 2z) = 2 + 2z$ times.

Consider the elements that occur once in $B_1, B_2, \dots, B_n$:
$b_1, b_2, \dots, b_t$. Thus, there are at most $\frac{2+2z}{2} = 1+z$ subsets in $B_1, B_2, \dots, B_n$ that do not contain the elements $b_1, b_2, \dots, b_t$. So, there exist $n - (z+1) = n-z-1$ subsets containing at least the elements $b_1, b_2, \dots, b_t$.

Suppose that $B_1, B_2, \dots, B_{n-1-z}$ contain the elements $\tilde{b}_1, \tilde{b}_2, \dots, \tilde{b}_{n-1-z}$ of $b_1, b_2, \dots, b_t$, respectively. Since
$$2(n - 1 - z) + z = 2n - 2 - z < 2n - 1,$$
there must exist an element $d$ that is not in $B_1, B_2, \dots, B_{n-1-z}$ but is in $B_{n-z}, \dots, B_n$.

Write $Y = \{a_1, \dots, a_z, \tilde{b}_1, \tilde{b}_2, \dots, \tilde{b}_{n-1-z}, d\}$, as desired.