Show that if $(a_k)_{k \in \mathbb{N}}, (b_k)_{k \in \mathbb{N}}$ are bounded, then $\lim \sup(a_k + b_k) \leq \lim \sup(a_k)+\lim \sup (b_k)$
I'm not quite sure how to prove this.
I want to try this comparing cases.
Assume that $a_k \geq 0 \forall k \in \mathbb{N}, b_k \geq 0 \forall k \in \mathbb{N}$
$\Rightarrow \sup (\bigcup_{i=k}^\infty \{a_k + b_k \}) = \sup (\bigcup_{i=k}^\infty \{a_k \}) + \sup (\bigcup_{i=k}^\infty \{a_k \})$
The same is true if both sequences are negative for all k.
If one of the sequences is negative for all indices and the other is positive, then the inequality would be real.
Would that be sufficient for a proof, or does the fact that e.g. all terms are positive in both sequences not imply the equality of the sup terms?
Your equality is not true, even if $a_k \geq 0$ and $b_k \geq 0$ for all $k$.
As a counterexample consider $a_k = 1 + (-1)^k$ and $b_k = 1 + (-1)^{k+1}$. Then $a_k + b_k = 2$ for all $k$, but $\sup_{i=k,\dots,\infty} a_i = 2$ and $\sup_{i=k,\dots,\infty} b_i = 2$ likewise.
But note that for arbitrary $i = k,\dots, \infty$ we have
$$ a_i + b_i \leq a_i + \sup_{\ell=k,\dots,\infty} b_\ell \leq \sup_{j=k,\dots,\infty}a_j + \sup_{\ell=k,\dots,\infty} b_\ell. $$
This should allow you to conclude the proof.
BTW: Note that you always wrote $\bigcup_{i=k}^\infty a_k$ instead of $\bigcup_{i=k}^\infty a_i$.