Conditinal probability theorem

I have to either prove or disprove the following:

If $\displaystyle P(A) \leq P(B)$, then $\displaystyle P(A|C) \leq P(B|C)$ (Assuming $\displaystyle P(C) \geq 0$)

After trying to prove this generally a million times and failing, I decided to find a counter example, I'm not sure if this works though.

All the following events have to do with rolling one fair dice.

A is getting a "2"

B is getting an odd number

C is getting an even number

Now $\displaystyle P(A) \leq P(B) $

i.e. $\displaystyle \frac{1}{6} \leq \frac{3}{6}$

but I believe that $\displaystyle P(B|C) = 0$

i.e $\displaystyle P( B \cap C) = 0$

and $\displaystyle P(A|C) = \frac{1/6}{3/6} = \frac{1}{3}$ (i.e $\displaystyle P(A \cap C) = \frac{1}{6}$

So this contradicts the hypothesis, but I am not sure if I calculated the probabilities of $\displaystyle P(A \cap C)$ and $\displaystyle P(B \cap C)$ right.

Thank you for your help.