Prove NP-completeness for union of NP-complete language and language in P
If for any two disjoint languages $X,Y$, we have:

$\textbf{*} \hspace{1mm} X \text{ is NP-complete} \land Y\in P \Rightarrow X\cup Y\text{ is NP complete}$ 

then $P\neq NP$.

Let $L$ be some NP-complete language. $L\cup\overline{L}=\Sigma^*$ is not NP-complete, thus, using (*) we obtain $\overline{L}\notin P$, but $L\in P\iff \overline{L}\in P$.

In fact your statement is equivalent to $P\neq NP$. You can prove the opposite direction by using the fact that if $P\neq NP$ and $L$ is NP-complete, then $\overline{L}$ is not in $P$. In that case, you can reduce $X$ to $X\cup Y$ by checking if the input is in $Y$, and output some constant word outside of $X\cup Y$ if it is. This can be done since we know now that $Y\subsetneq \overline{X}$.