Step 1: Define reducing character.
A reducing agent donates electrons and itself gets oxidized. A compound shows reducing character when its central atom is below its maximum oxidation state and can still be oxidized further.
Step 2: Find the oxidation state of S in $SO_2$.
\[ x + 2(-2) = 0 \Rightarrow x = +4 \] Sulfur in $+4$ can be further oxidized to $+6$ (as in $SO_3$ or $SO_4^{2-}$). So $SO_2$ acts as a reducing agent.
Step 3: Find the oxidation state of S in $SO_3$.
\[ x + 3(-2) = 0 \Rightarrow x = +6 \] This is the maximum oxidation state of sulfur. $SO_3$ cannot be easily oxidized further, so it does not show reducing character.
Step 4: Analyze $TeO_2$.
In $TeO_2$: $x = +4$. Tellurium is in a lower oxidation state, but $Te^{+4}$ compounds are much less effective reducing agents than $S^{+4}$ compounds under standard conditions.
Step 5: Analyze $TeO_3$.
In $TeO_3$: $x = +6$. Tellurium is at its highest oxidation state here, so $TeO_3$ shows no reducing character.
Step 6: Compare and conclude.
$SO_2$ is the well-established reducing agent among these four compounds. It reduces permanganate, dichromate, and halogens under standard conditions.
Step 7: State the final answer.
\[ \boxed{SO_2} \]