Prove that subsequence converges to limsup

Since $\alpha=\limsup x_n$, by the definition of $\limsup$, there is some $x_{n_1}$ with $|x_{n_1}-\alpha|<{1\over 2}$. (That's the crucial step, so be sure you understand why.)

Similarly, there is some $x_{n_2}$ with $|x_{n_2}-\alpha|<{1\over 2^2}$. Continuing, for each $k\in\mathbb{N}$, there is some $x_{n_k}$ with $|x_{n_k}-\alpha|<{1\over 2^k}$.

Then $\{x_{n_k}\}\subset \{x_n\}$ and $x_{n_k}\to \alpha$ as $k\to\infty$.