Which element is being reduced in this reaction?
Your reasoning is not entirely right. In those compounds, O is always (-II), K and Na always (+I). However, it's not true for S, which can have different oxidation states. So does Mn.

So, in KMnO4, Mn must be (+VII). In MnO2, it is (+IV). Thus, Mn is reduced.

For sulfur: in NaSO3, it must be be (+V). In NaSO4, it must be (+VII). Thus, sulfur is oxidized.

However (big however, in fact): the sulfate anion is not SO4–, but SO42–. Similarly, sulfite is SO32–, not SO3–. So, your equation should be instead:

$$\ce{2KMnO_4 +3 Na_2SO_3 +H_2O -&gt; 2MnO_2 +3Na_2SO_4 +2KOH}$$

which means that sulfur is oxidized from (+IV) to (+VI).