Which element is being reduced in this reaction? Which element is reduced in this reaction:
  $$\ce{2KMnO_4 +3NaSO_3 +H_2O -&gt;2MnO_2 +3NaSO_4 +2KOH}$$


I am trying to figure out which elements lose their electrons, but how am I supposed to know when elements are neutral and when they are ionized? I have tried the following, but I don't believe it to be correct

$$\ce{2K^+Mn^+O_4^{-2} +3Na^+SO_3^{-2} +H_2O -&gt;2Mn^{+2}O_2^{-2} +3Na^{+}SO_4{-2} +2K^+OH^-}$$

My reasoning being that:


Oxygen atoms are always doubly anionized 
Sulfate and sulfite are also doubly ionized
Sodium and potassium are always singly cationized


Based on these rules I tried to make the bonds neutral, but my assumptions must be incorrect. Any help would be greatly appreciated
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).