Explanation of calculating equilibrium constant
Your reasoning is correct. Initially, you saw that there was no iodine in the vessel. After reaching equilibrium, you saw 0.25 mol. Where did this come from? 

Answer: Hydrogen iodide broke down to form this iodine and hydrogen. How much hydrogen iodide? Well this is a super simple moles problem. 

0.25 moles of iodine was formed, and the mole ratio for $\ce{HI:I_2}$ is $2:1$. For 0.25 moles of iodine to be formed, 0.5 moles of hydrogen iodide must have been used (just look at your balanced equation).

If 0.5 moles of hydrogen iodide was used, how much hydrogen iodide is left? 0.60 moles minus 0.50 moles to leave 0.10 mol. 

Then, as you did correctly, you plug it into the $K_c$ formula. 



While you have stated that you just want to know how to do it (which you should be able to now), it is useful to know what's going on. Basically, the $K_c$ is a measure of where the equilibrium is. If your $K_c$ value is high, there will be a higher concentration of the products at equilibrium.