Effect of Temperature on pH of Water
TL;DR
$\rm pH=6$ is the neutral pH at $\rm 50\,^{\circ} C.$
Long Version
Actually there is nothing wrong with your calculations. It seems that you are confused about the definition of a solution being acidic. If something is acidic, this means that the concentration of $\ce{H+}$ ions (technically $\ce{H3O+}$ ions) is greater than the concentration of $\ce{OH-}$ ions. At room temperature (25 degrees Celsius) a solution is neutral if its pH is 7. This means that for a solution to acidic, its pH must be lower than 7 at 25 degrees.
When we increase the temperature to 50 degrees, the pH of a neutral solution is actually 6, not 7. You can check this as if you were find the total concentration of $\ce{H+}$ ions and $\ce{OH-}$ ions, you will find that they both equal $10^{-6}$. Since there concentration is equal, the solution must be neutral. So for a solution to be acidic at a temperature of 50 degrees, its pH must be lower than 6, not 7.
Explanation
Now you might wonder what the reason for this. Consider the following equation which is the auto-disassociation of water $$\ce{2H2O + heat \leftrightharpoons H3O+ + OH-}$$ As you can see, this reaction is actually endothermic. This means that when you increase the temperature, due to Le Chatelier's principle, the equilibrium will shift to the right; hence the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions will increase. This is indicated by the increase in the $\rm K_w$ value, as more products are formed.
Therefore since there is a greater of concentration of $\ce{H3O+}$, a solution which is neutral will have more $\ce{H3O+}$ ions at 50 degrees than it would when it is 25 degrees. Therefore its pH will be lower than 7 despite being neutral.