If neither $L_1$ nor $L_2$ are context free then is $L_1 \cup L_2$ also not a context free language?
First, note that CFL is not closed under intersection, contrary to what you've mentioned.
For example:
$$ L_1 = \{a^ib^ic^j|i,j \geq 0\}, \quad L_2 = \{a^ib^jc^j|i,j\geq 0\}. $$
We see that $L_1 \cap L_2 = \{a^ib^ic^i|i\geq0\}$,
which is not CFL.

For your question,
Take two languages $L_1, L_2$ which are not CFL.
Let
$$
\begin{align*}
L'_1 &amp;= \{0x|x \in L_1 \} \cup \{1x|x \in \Sigma^*\}, \\
L'_2 &amp;= \{1x|x \in L_2 \} \cup \{0x|x \in \Sigma^*\}, \\
\end{align*}
$$
which are both not CFL. It's easy to see that $L'_1 \cup L'_2 = \Sigma^*$, which is CFL (as it is a regular language).