Why is the empty set finite?

That definition is a (rare) example of Rudin doing things inefficiently. He could have defined $J_n$ for each non-negative integer $n$ to be the set of non-negative integers less than $n$, so that $J_0=\varnothing$, $J_1=\{0\}$, $J_2=\{0,1\}$, etc. Then he could have defined a set $A$ to be finite if and only if $A\sim J_n$ for some $n\in\Bbb N$ (where $\Bbb N$ includes $0$). This is essentially the usual set-theoretic definition stripped of some set-theoretic detail that would be out of place here.

The parenthetical remark is just saying that formally the definition of finite set does not apply to the empty set, but the empty set is taken to be finite by convention.

If this bothers you, note it is possible to define a set as infinite precisely when there exists a proper subset of it with a bijection to the set. This captures in one go what it means to be infinite.

What they are saying here is that the empty set is considered to be finite by convention, or for a different reason. If they were being more precise, they would have said that a set is finite if it is empty or is in bijection with some $J_n$.

Suppose in general that for non-negative integers $n$ we defined $J_n$ to be the set of positive integers no greater than $n$. This certainly matches the definition given for $J_n$ when $n\ge 1$, but what about when $n=0$? Well, there aren't any positive integers no greater than $0$, so $J_0=\emptyset$! Certainly the empty set is in bijection with itself. In that way, we may extend the definition of finite, without relying on a "just roll with it" in the case of the empty set.