If neither $L_1$ nor $L_2$ are context free then is $L_1 \cup L_2$ also not a context free language? If two regular languages $L_1$ and $L_2$ are both not context free languages then is $L_1 \cup L_2$ also not a context free?

I am aware that if $L_1$ and $L_2$ are context free languages then the language $L_1 \cap L_2$ is also context free but cannot quite connect the dots. If someone could help out that would be great.
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).