How to identify a redox reaction? Which balanced equation represents a redox reaction?
  \begin{align}
&amp;\mathrm{A.} &amp;\ce{AgNO3 + NaCl &amp;-&gt; AgCl + NaNO3} \\
&amp;\mathrm{B.} &amp;\ce{BaCl2 + K2CO3 &amp;-&gt; BaCO3 + 2KCl} \\
&amp;\mathrm{C.} &amp;\ce{CuO + CO &amp;-&gt; Cu + CO2} \\
&amp;\mathrm{D.} &amp;\ce{HCl + KOH &amp;-&gt; KCl + H2O}
\end{align}


I don't understand how to identify a redox reaction. The answer key shows the correct answer as D, but I'm confused as to why.

D shows an acid-base reaction. Are all acid-base reactions redox reactions and why aren't the other answer choices correct?
You are confused because the answer key is wrong.
The way to figure it out is to assign oxidation numbers to each species following the rules. Oxidation states are simply a convenient way to keep track of electrons without having to draw out Lewis structures, count electrons pairs and so on. The rules help make that happen.

Notice the position of the sign relative to when you indicate charge. It is not the same.

The correct answer is C though: $$\ce{CuO + CO -&gt; Cu + CO2}$$
with oxidation states 
$$\ce{\overset{(+2)}{Cu}\overset{(-2)}{O} 
    + \overset{(+2)}{C}\overset{(-2)}{O} 
   -&gt; \overset{(0)}{Cu} 
    + \overset{(+4)}{C}\overset{(2\times(-2))}{O2}}$$
Notice Cu went from +2 to 0. It gained 2 electrons to do that, thus it was reduced. What gave up the electrons? The other carbon, because it went from +2 to +4: its oxidation state increased by 2, thus it lost 2 electrons.

Recall OIL RIG: Oxidation is loss of electrons and reduction is gain of electrons. 
Whatever is oxidized increases its oxidation state, and whatever is reduced decreases it.