I am trying to prove that if $x_n$ is a sequence such that every subsequence $a_n$ has a subsubsequence $b_n$ whose $\limsup b_n \le M $ then $\limsup x_n \le M $
if I take as a subequence $a_N = sup_{n>N} x_n$ then it is monotone decreasing, so that $sup_{n>N} a_n = a_N $ therefore $\limsup a_n = inf_{N} sup_{n>N} a_n = inf_{N} a_N = inf_{N} sup_{n>N} x_n = \limsup x_n $
$b_k$ is a subsequence of $a_n$, so $\limsup a_n \ge \limsup b_k $, and since $a_n$ is monotone decreasing so is $b_k$, hence $\limsup b_k = inf_{K} sup_{k>K} b_k = inf_{K} b_K = inf_{K} a_{n_k} \ge inf_N a_N = \limsup a_n $
so $\limsup b_k = \limsup a_n = \limsup x_n $ and the result follows
I am not too sure of my demonstration, since I have not done this for some years... Are those correct statements ?