If $\{E_n\}_{n\ge1}$, and $\{F_n\}_{n\ge1}$ are increasing and independent for each $n$, show that their limits are independent.
Here is my attempt:
Note that $\{E_n \cap F_n\}$ is also increasing. Then, \begin{align} P(E \cap F) &=P\left( \lim_{n \to \infty}\{ E_n\} \cap \lim_{n \to \infty}\{ F_n\} \right)\\ &=P\left( \bigcup_{n=1}^{\infty} \{ E_n \}\cap \bigcup_{n=1}^{\infty} \{ F_n \} \right)\\ &= P\left( \bigcup_{n=1}^{\infty} \{ E_n \cap F_n\} \right) \\ &= P(\lim_{n \to\infty} \{E_n \cap F_n \}) \\ &=\lim_{n\to\infty} P(E_n \cap F_n )\\ &=\lim_{n\to\infty}P(E_n)P(F_n)\\ &=P(E)P(F) \hspace{5mm} \blacksquare \end{align}
What you did is essentially correct. Here are some remarks.
It is not necessary to write the $\lim$ of a sequence of sets (unless it has been properly defined with the equality of $\limsup$ and $\liminf$ of the corresponding sequence). All we actually need is that if $\left(A_n\right)_{n\geqslant 1}$ is a non-decreasing sequence of sets, then $\lim_{n\to +\infty}\mathbb P\left(A_n\right)=\mathbb P\left(\bigcup_{j=1}^{+\infty}A_j\right)$.
The place where the assumption that the sequences $\left(E_n\right)_{n\geqslant 1}$ and $\left(F_n\right)_{n\geqslant 1}$ are non-decreasing is used should appear more clearly. It is used to justify the equality
$$ \left( \bigcup_{j=1}^{+\infty}E_j\right)\cap \left( \bigcup_{k=1}^{+\infty}F_k\right)= \bigcup_{n=1}^{+\infty}\left(E_n\cap F_n\right). $$ The last equality is not true in general (take $E_{2i}=\emptyset=F_{2i+1}$, $E_{2i+1}=F_{2i}=\Omega$ for example.