Union of non-context free languages
Yes, and this works for regularity and decidability as well! (not that this is very helpful in any sense)
For example, take any problem $A$ such that $A$ and its complement $\overline{A}$ are not CFLs (the halting problem as an extreme example)
Their union is $\Sigma^*$ which is obviously a CFL.

As a side note, and thanks for @D. Ben Knoble for pointing this out in the comments, that this is only an example. It does not necessarily hold for any two non-CFL languages, and on the contrary also a union of two CFL's may (not always!) end up not being a CFL.