The multiplication rule for independent events

Step 1 (set up the counting): let $|A| = a$ and $|B| = b$. An outcome of $\Omega = \Omega_1 \times \Omega_2$ satisfying both conditions is a pair $(\omega_1, \omega_2)$ with $\omega_1 \in A$ and $\omega_2 \in B$.

Step 2 (apply the multiplication rule for counting): choosing $\omega_1$ can be done in $a$ ways and, independently of that choice, $\omega_2$ can be done in $b$ ways, so by the multiplication rule for counting there are $a \cdot b$ favorable pairs.

Step 3 (divide by the total number of outcomes): the total sample space has $m \cdot n$ equally likely outcomes, so $P(A \cap B) = \frac{ab}{mn}$.

Step 4 (factor the fraction): rewriting $\frac{ab}{mn} = \frac{a}{m} \cdot \frac{b}{n} = P(A) \cdot P(B)$, since $P(A) = a/m$ and $P(B) = b/n$ by the classical probability formula, which proves the claim.