"Definition of NP via relations and quantifiers; not via NTMs"
I think I just realized, after finding a similar problem online, that this is asking to simply show that ${\sf NP}$ is closed under union. Can anybody give any insight into whether or not this is a proper solution, or if I'm missing anything? I mostly understand my answer, but I don't really know what I'm doing with the $c$ and $d$ constants... I put them in there because I feel like I'm going to need to.


  This question is asking to prove that ${\sf NP}$ is closed under union. It is the same as asking if $L=L_{1}\cup L_{2}$ is in ${\sf NP}$.
  
  Since $L_{1}$ and $L_{2}$ are both in ${\sf NP}$, it follows that they both have polynomial time verifiers $R_i(x_i,y_i)$ that evaluate to true when $y_{i}$ is a valid certificate/witness for $x_i\in L_i$, where $i\in \{1,2\}$.
  
  To show that $L=L_1\cup L_2$ is also in ${\sf NP}$, we need to construct a polynomial-time verifier $R(x,y)$. The certificate $y=\langle y_1,y_2 \rangle$ for $L$ will be an encoding of the two certificates $y_1$ and $y_2$ in such a way that $R(x,y)$ evaluates to true when $\displaystyle\bigvee_{1\leq i\leq 2} R_i(x,y_i)$ evaluates to true.
  
  Given that $R_{i}(x_i,y_i)$ for $i\in\{1,2\}$ are polynomial time verifiers with running time $O(c_i|x_i|^{d_i})$ for some $c,d\in\mathbb{N}$, $R(x,y)$ will have a running time of at most $O(2\cdot max(c_i)|x|^{max(d_i)})$, which is still polynomial. Therefore, $L\in {\sf NP}$.