$\limsup$ and cluster points

Put $L = \limsup_n x_n$ and $s_n = \sup_{k \ge n} x_k$. Notice that $s_n$ is a decreasing sequence in $n$ and that $s_n \downarrow L$.

Let $\lambda > L$. then for some $n$, $\sup_{k\ge n} x_k <\lambda$. Hence there can be no limit point of the sequence in $(\lambda, \infty)$. Since $\lambda > L$ was chosen arbitrarily, all limit points of the sequence must lie in $(-\infty, L]$.

We will be done if we can show that the limsup is a limit point of the $x_n$. We choose a subsequence as follows. Choose $n$ so that $\sup_{k\ge n} x_k < L - 1/2$; now fix $n_1$ so $\sup_{k\ge n_1} x_k > L - 1/2$. Now suppose $n_1 < n_2 < \cdots < n_k$ are chosen. Since $s_n\downarrow L$ there is $N$ so that $s_N > L - 1/2^n$. Since $S$ is decreasing, we can choose $N > n_k$. Put $N = n_{k+1}$.

The sequence $\{x_{n_k}\}$ converges to $L$, so $L$ is a limit point of the $x_n$.