Note that for two sequences $a_n$ and $b_n$, it is always true that:
$\sup (a_k+b_k) \leq \sup (a_k)+ \sup(b_k)$
Just by $\{ a_k+b_k :k\geq n \}\subseteq \{ a_k+b_l: k,l\geq n \}$
So you can conclude that $\sup_{k\geq n}(x_k+y_k)\leq \sup_{k\geq n}x_k+\sup_{k\geq n}y_k$.
Denote $c_n:=\underset{k\geq n}{\inf} a_k$ and $d_n:=\underset{k\geq n}{\inf} b_k$
Assuming that $\infty>\inf c_n>-\infty$ and $\infty>\inf d_n>-\infty$, for all $\epsilon>0$ there exists $i$ and $j$ such that:
$d_j<\inf d_n+\frac{\epsilon}{2}$ and $c_i\leq \inf c_n+\frac{\epsilon}{2}$. Then:
$\inf(c_n+d_n)\leq c_i+d_j \leq \inf c_n+ \inf d_n +\epsilon$
Since this is true for all $\epsilon>0$, this must mean that:
$\inf(c_n+d_n)\leq \inf c_n+ \inf d_n$, and thus:
$\inf_n\sup_{k\geq n}(x_k+y_k)\leq \inf_n\Big(\sup_{k\geq n}x_k+ \sup_{k\geq n}y_k \Big) \leq\inf_n\sup_{k\geq n}x_k+\inf_n\sup_{k\geq n}y_k$
This looks valid to me! You show that the $\lim \limits_{n \to \infty}$sup{${x_n, x_{n+1},...}$} is equal to sup(T). This shows the two definitions are equal.
Your steps look good. You show that $\beta \leq \alpha$ by establishing a cutoff point after which $x_{n_k} \leq\space $sup{$x_k, x_{k+1}, ...$}. You then take the limit of each side of this inequality. This should be valid since the left side is less than the right for all $x_{n_k}$.
You then show that $\alpha \leq \beta$ by constructing a subsequence of $x_n$ that converges to $\alpha$ [which establishes $\alpha$ as a cluster point of $(x_n)$] and then using the fact that $\beta$ is the supremum of all cluster points of $(x_n)$.
I was unable to find any flaws in these steps.
Best Answer
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 $x_n$.