This proof will mostly rely on using definitions.

We begin by recalling the definition of independence. Events $A$ and $B$ are independent if and only if $P(A\cap B) = P(A)P(B)$. Translating definition to the current case, we are interested in evaluating whether $P(A \cap \emptyset) = P(A)P(\emptyset)$.

First consider $P(A \cap \emptyset)$. By monotonicity, $P(A \cap \emptyset) \le P(\emptyset)$. From the async, we know that $P(\emptyset) = 0$. Since probabilities are non-negative, $P(A \cap \emptyset) = 0$.

For the right hand side, we know again that $P(\emptyset) = 0$ so $P(A)P(\emptyset)=P(A) \cdot 0 = 0$, so both sides of the equation are zero. This means that $\emptyset$ is independent of $A$