I was working on a problem(picture) and it got me wondering about the cases in which the equality doesn't hold. It got me to come up with a theorem. I'd like to know if this is true (I think it is) and a proof would be nice(working on one, but not complete yet) I basically defined K_S and K_T to be "indices" in which the respective limsups appear.
$Theorem:Let\ S_n\ and\ T_n\ be\ both\ bounded\ but\ both\ non convergent\ sequences. \\Let\ K_S := \{n|n\in\mathbb{N}\ and\ S_n = lim sup (S_n)\} \\Let\ K_T := \{n|n\in\mathbb{N}\ and\ T_n = lim sup (T_n)\} \\If K_S \cap K_T = \varnothing\ ,\ then\ \\lim sup (S_n+T_n) < lim sup (S_n)+lim sup (T_n)$
I understand your idea but here is the thing. You have bounded sequences, hence by Bolzano-Weierstrass theorem we have that there exists an increasing subsequence of $T_n$ which converges to the $limsupT_n$. Same for $S_n$. So even if these set have an empty intersection, what you get at best is the well known inequality $$\limsup(a_n+b_n)\leq \limsup a_n+\limsup b_n.$$
I want to note though that your effort and your energy put into your proof will not be fruitless just because you did not prove something new. You are doing fine, and these little efforts will improve you. This is what mathematics are for.