Sulphur dioxide is oxidised in air by a mechanism with two steps…

Step 1. Add the steps. Double step 1: $2\text{SO}_2 + 2\text{NO}_2 \rightarrow 2\text{SO}_3 + 2\text{NO}$. Add step 2: $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$. The $\text{NO}$ and the $\text{NO}_2$ cancel. What remains is $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$.

Step 2. The catalyst. $\text{NO}_2$ is there at the start, used in step 1 and made again in step 2. It is the catalyst.

Step 3. The intermediate. $\text{NO}$ is made in step 1 and used up in step 2. It is the intermediate.

Step 4. The rate law. The slow step sets the rate, and it is a collision of $\text{SO}_2$ with $\text{NO}_2$. So $\text{rate} = k[\text{SO}_2][\text{NO}_2]$. The catalyst is in the rate law though not in the overall equation.