I will prove this using my favourite definition of the Limit Superior and Limit Inferior. Drop a comment if you need more help.
Let $ \mathscr L_{(x_n)} = \{l \in \Bbb R \ | \ \text{there is a subsequence of $ (x_n) $ which converges to $l$}\}$
We will prove that $ \mathscr L_{(x_n)} = \{ x, y \} $.
Well one direction is easy since we know that there are two subsequences of $(x_n)$ - namely, $(x_{2n})$ and $x_{(2n + 1)}$, respectively - which converge to $ x $ and $y$. Hence, $ \{ x, y \} \subseteq \mathscr L_{(x_n)} $.
Now prove that there is nothing in $ \mathscr L_{(x_n)} $ other than $x$ and $y$. To this end, suppose there exists $ p \in \mathscr L_{(x_n)} $ such that $ p \not \in \{x, y\} $. Then there is, by definition, a subsequence, $(x_{n_i}) $ of $(x_n)$ such that $ \lim \limits_{i \to \infty } x_{n_i} = p $. Now let $\epsilon = \min \{ |p - x|, |p - y| \}$. Then there is $I \in \Bbb N$ such that $ i \gt I \implies | x_{n_i} - p | \lt \frac{ \epsilon}{2} $. But this would mean that for $i \gt I$,
$$ \frac{\epsilon}{2} = \epsilon - \frac{\epsilon}{2} \le | x - p |- \frac{\epsilon}{2} \lt | x - p | - | x_{n_i} - p | \le | x - p - x_{n_i} + p | = |x_{n_i} - x| $$
and this contradicts the fact that $(x_{2n}) \to x$.
Hence, $$ \mathscr L_{(x_n)} = \{ x, y \} $$
Now if you will grant me that a finite set has a maximum and a minimum then,
$$ \lim \sup x_n = \sup \mathscr L_{(x_n)} = \max \mathscr L_{(x_n)} = \max \{x, y\} \;\; \text{and} $$
$$ \lim \inf x_n = \inf \mathscr L_{(x_n)} = \min \mathscr L_{(x_n)} = \min \{x, y\} $$