The short answer is: Yes, it is true, because the contrapositive just expresses the fact that $1^2=1$.
But in controversial discussions of these issues, it is often (but not always) a good idea to try out non-mathematical examples:
"If a nuclear bomb drops on the school building, you die."
"Hey, but you die, too."
"That doesn't help you much, though, so it is still true that you die."
"Oh no, if the supermarket is not open, I cannot buy chocolate chips cookies."
"You cannot buy milk and bread, either!"
"Yes, but I prefer to concentrate on the major consequences."
"If you sign this contract, you get a free pen."
"Hey, you didn't tell me that you get all my money."
"You didn't ask."
Non-mathematical examples also explain the psychology behind your teacher's and classmates' thinking. In real-life, the choice of consequences is usually a loaded message and can amount to a lie by omission. So, there is this lingering suspicion that the original statement suppresses information on 0 on purpose.
I suggest that you learn about some nonintuitive probability results and make bets with your teacher.
You are right when you say that $\forall x$, the statement $x \in \emptyset$ is false. This means that $\forall x, \neg (x \in \emptyset)$, which is equivalent to $\forall x, x \notin \emptyset$. Perfectly true.
Then you say "the statement $\forall x \in \emptyset$ is false". $\forall x \in \emptyset$ is NOT a statement, it's an incomplete sentence. Either you write "$\forall x, P(x)$", either you write "$\forall x \in X, P(x)$", which is a shorthand for "$\forall x, (x \in X \implies P(x))$". "$\forall x \in \emptyset$" is not a statement. It can't be true or false.
$\forall x \in \emptyset, P(x)$ is a shorthand for $\forall x, (x \in \emptyset \implies P(x))$, which is equivalent (since $x \in \emptyset$ is always false) to $\forall x, ~\textrm{false} \implies P(x)$. After looking at the truth table for $\implies$, this is equivalent to $\forall x, ~\textrm{true}$ (whatever $P(x)$ may be), which is $\textrm{true}\;.$
If you want to disprove $\forall x \in \emptyset, P(x)$ you have to show me an $x \in \emptyset$ such that $P(x)$ is false. Well you will never find an $x \in \emptyset$.
Best Answer
This statement is false.
Take $ a=\frac{\epsilon}{2}$.
we have $ |a|\le \epsilon $ but $ a\ne 0 $.
to be true, you should change the position of the parenthesis :
$$\forall a\in \Bbb R \;\; (\forall \epsilon>0 \;\; |a|\le \epsilon \implies a=0)$$
to prove it is true, assume that $ a\ne 0 $.
then there exist $ \epsilon =\frac{|a|}{2} $ such that
$$|a|>\epsilon$$