Let us assign oxidation numbers to each species in the given reaction:

$$+1 \quad -2 \quad \quad +1 \quad -2 \quad \quad 0 \quad \quad +4 \quad -2$$ $$2\text{Cu}_2\text{O(s)} + \text{Cu}_2\text{S(s)} \rightarrow 6\text{Cu(s)} + \text{SO}_2(\text{g})$$

In this reaction, copper is reduced from the +1 state to zero oxidation state, while sulphur is oxidised from the -2 state to the +4 state. Since both oxidation and reduction are occurring simultaneously, the above reaction is a redox reaction.

Further, $\text{Cu}_2\text{O}$ helps sulphur in $\text{Cu}_2\text{S}$ to increase its oxidation number, therefore, Cu(I) is an oxidant. The sulphur of $\text{Cu}_2\text{S}$ helps copper both in $\text{Cu}_2\text{S}$ itself and $\text{Cu}_2\text{O}$ to decrease its oxidation number; therefore, sulphur of $\text{Cu}_2\text{S}$ is the reductant.