You're starting your indirect proof wrong. What you want to prove is
A: For every choice of 64 days, at least one day-of-the-week will be hit more than 9 times.
In order to conduct an indirect proof, you start by assuming the negation of (A). However that negation is not, as you write
B: if you choose 64 days at random, then, at most, nine of those days will be the same day of the week.
There's nothing about randomness in A, and negating a claim that's not about randomness does not make it into one. So if nothing else, you're making your task harder here by suddenly introducing randomness.
However, the negation of A is not even
C: For every choice of 64 days, at most nine of those days will be the same day of the week.
C is quite obviously false because nothing stops me from choosing 64 Thursdays, which will then be a counterexample. If C and A had been negations, then this would constitute an indirect proof of A -- but clearly and intuitively just because I can choose 64 Thursdays doesn't mean that every choice I can make must contain 10 of some day. (And indeed if the argument was valid, it would be hard to argue what would be wrong with the same proof with "63" instead of "9").
The actual negation of (A) is
D: There is at least one way to choose 64 days such that each day of the week is used at most 9 times.
If we assume that, we get something concrete to work with, namely a set $S$ of 64 days where each weekday appears at most 9 times. We can then begin to construct a contradiction, such as by verifying that the size of $S$ is 64 by adding the number of Mondays in the set to the number of Tuesdays, Wednesdays, and so on.