Intersection with a finite set
There is no need to consider the intersection $L \cap F$. If $F$ is finite, then so is $L \cap F$, but that is not a useful fact here.

Insteed, one should proceed by contradiction. Suppose $L \setminus F$ were a context-free language. Then there exists a context-free grammar $G_1 = (V_1,\Sigma,R_1,S_1)$ such that $L(G_1) = L \setminus F$. But since $F$ is finite, then $F$ is trivially a context-free language. Let $G_2 = (V_2,\Sigma,R_2,S_2)$ be a context-free grammar such that $L(G_2) = F$.

But then $G = (V,\Sigma,R,S)$ where $S \notin V_1 \cup V_2$ and

$$V = V_1 \cup V_2 \cup \{ S \}$$
$$R = R_1 \cup R_2 \cup \{ S \rightarrow S_1, S \rightarrow S_2 \}$$

is a context-free grammar such that $L(G) = L$. This is a contradiction.