I know this is clearly false, but I was given a "proof" by contradiction that seems to make a case. Why does it fail?
Begin by assuming that all natural numbers cannot be divisible by $9999$
Let $A = \{a \in \mathbb{N} | a$ is not divisble by $9999 \}$ represent all the counterexamples. We know that $A \neq \emptyset$ and must contain a least element of $b$.
Since $0$ is divisible by $9999$ we can conclude that $0 \not \in A$ and therefore $b \neq 0$. Now for an example $b - 9999$ where we know that $b - 9999 < b$. Therefore, we know that $b - 9999$ is not a valid counterexample and cannot be in $A$. Therefore, $b - 9999$ is divisible by $9999$ such that there's an integer $c$ where $b - 9999 = 9999c$.
We then get $b = 9999 + 9999c = 9999(1+c)$ and prove that $b$ is divisible by $9999$, contradicting $b \in A$.
$\mathbb{N}$ is not closed under subtraction. For any natural number $n$ and any natural number $m \geq n$, subtracting another natural number $n-m$ gives a number that is not a natural number. So, in order for $b-9999$ to be in $A$ at all, as $A$ is restricted to only natural numbers, they have to first show that $b \geq 9999$.
Also, even if the rest of the proof was somehow correct, they negated their quantifier at the beginning incorrectly. The negation of "for all $a$, $p$ is true" is not "for no $a$, $p$ is is true", but "there exists at least one $a$ such that $p$ is false. So, in this proof, if you are trying to "prove" that all natural numbers are divisible by $9999$, then if using contradiction, you would assume that there exists at least one natural number such that it is not divisible by $9999$.