How is this a good proof 

To me this just reads: We suppose this and that and because we supposed it, it must be true.
Why does this show that $H$ is closed under inverse? Why does it show that $H$ is closed under the operation? It doesn't, it just assumes it.
Am I misunderstanding this?
 A: Read carefully for the hypotheses and the conclusion:
Hypotheses: $H$ is a subset of $G$, $H$ is closed under the operation, $H$ is closed under inverses.
Conclusion: Then $H$ is a subgroup of $G$.
Normally, given a random subset $H$ of $G$, we would have to check that it is closed under the operation, that it is closed under inverses, and that the identity is in $H$. (or maybe your book has a slightly different definition of subgroup, but the main point is, the definition requires you to check a (longish) list of properties)
This theorem is saying, in essence, that if the first two properties are satisfied, the third will be automatically satisfied (i.e. there is no need to verify that whole long list from the definition). The referred-to theorem (3.1) reduces this to showing $ab^{-1} \in H$. (i.e. {closed under inverses} and {closed under operation} $\Longrightarrow$ $ab^{-1} \in H \Longrightarrow H$ is a subgroup)
Explanation of Theorem 3.1:
Here, we want to show that satisfying the hypothesis "$ab^{-1} \in H$ whenever $a,b\in H$" is enough to conclude $H$ is a subgroup. That is, in the future, if we can simply check $ab^{-1} \in H$, there is no need to verify the usual requirements of a subgroup. (looks like this list is "operation is associative", "closed under inverses", "closed under operation" and "identity is in the set")
Assuming you understand how the other properties are verified, we need to show the identity is in such a set. If we choose $a = x$, $b = x$, then we know $ab^{-1}$ is in $H$. But $ab^{-1}$ is nothing more than $xx^{-1} = e$, so that is verified.
Many of these beginning abstract algebra proofs will seem... tautological sometimes. (i.e. you may find yourself thinking "what is there to even prove?!?") Take it as an exercise in patience and careful understand of definitions (and how to verify them), and take comfort in the fact that things will become a lot less trivial in time.
A: It seems to me that the proof assumes that associativity is inherited (it doesn't prove that the operation on $H$ is associative, but it is associative in $H$ because associative in $G$), and all you really have to do to prove that $H$ is a subgroup is to show it contains the identity - because then it is closed under an associative binary operation with identity and inverses.
For this we need the condition that $H$ is nonempty, which isn't used explicitly in the proof as stated. Take $a\in H$ (which exists because $H$ is nonempty) then by the existence of inverses $a^{-1}\in H$ and then by closure under the binary operation $aa^{-1}=1\in H$.
