What do the half-reactions need to be for balancing a redox reaction in an acidic medium?
Below, the number mentioned in the parenthesis is the oxidation number of sulfur.


Writing down half reactions:
\begin{align}
\ce{CuS^{(-2)}     &amp;-&gt; ^{(4)}SO2}&amp;&amp; \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} &amp;-&gt; ^{(4)}SO2}&amp;&amp; \text{Reduction}
\end{align}
Adding electrons in both reactions:
\begin{align}
\ce{CuS^{(-2)}           &amp;-&gt; ^{(4)}SO2 + 6e-}&amp;&amp; \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- &amp;-&gt; ^{(4)}SO2}&amp;&amp; \text{Reduction}
\end{align}
Balancing $\ce{Cu}$:
\begin{align}
\ce{CuS^{(-2)}&amp;-&gt; ^{(4)}SO2 + 6e- + Cu^{+2}}&amp;&amp; \text{Oxidation}
\end{align}
Balancing charges of both half reactions:
\begin{align}
\ce{CuS^{(-2)}&amp;-&gt; ^{(4)}SO2 + 6e- +CuO + 6H+}&amp;&amp; \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &amp;-&gt; ^{(4)}SO2}&amp;&amp; \text{Reduction}
\end{align}
Balancing oxygen and hydrogen:
\begin{align}
\ce{CuS^{(-2)} + 3H2O&amp;-&gt; ^{(4)}SO2 + 6e- +CuO + 6H+}&amp;&amp; \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &amp;-&gt; ^{(4)}SO2 + 2H2O}&amp;&amp; \text{Reduction}
\end{align}
Adding 3 copies of reduction half reaction with oxidation half reaction, gives finally balanced redox reaction:
$$\ce{CuS + 3SO4^{2-}  + 6H+ -&gt; 4SO2 +CuO + 3H2O}$$


(Thanks to Silvio Levy for helping me to solve this problem.)