Prove that if H is infinite, then $\frac{1}{H}$ is infinitesimal Prove that if H is infinite, then $\frac{1}{H}$ is infinitesimal.
I try to do prove it this way:

But as I'm an amateur at proof-writing, with almost non-existent knowledge in proof-writing, my question is as follows:

Is this proof written correctly? If not, what have I done wrong/ could I have done better? (This is NOT homework). It may contain errors, I did not see them at the time, if so. Please show me these errors, if you would like to.

I ask this question as I want to become more certain in myself at writing proofs. Or, if I have made errors, learn what I did wrong.
 A: This is the usual rigorous proof in Internal Set Theory that given an unlimited real $u$, $1/u$ is infinitesimal. This uses the notation $\forall^\mathsf{s}$ which means "for all standard", where the standard reals are the reals uniquely definable without IST (so, the "usual" reals). The proof just follows from the usual definitions and definition chasing. If you have not seen these definitions before, you can see where they come from by staring for a bit.
Since $u$ is unlimited, we know by definition that $\forall^\mathsf{s}y>0,\lvert u\rvert>y$.
For a given $y>0$, this implies $\lvert 1/u\rvert<1/y$.
Given a standard $a>0$, pick $y=1/a>0$ so that $\lvert 1/u\rvert<a$.
We conclude $\forall^\mathsf{s}a>0,\lvert 1/u\rvert <a$.
This is exactly the statement $1/u$ is infinitesimal (by definition).

Regarding your proof: it's difficult to provide exact feedback without being familiar with all the peculiarities of Keisler's approach. However, you should start with the definition of what Keisler means by infinite and infinitesimal and attempt to relate them.
A: Your proof is very sketchy. 
Consider the set $S = \mathbb R - {0}$. Define a binary operation (*) on $S$. 
Since (*) is a binary operation from $S\times S \to S$, it satisfies:
1) $\forall a,b \in S, a*b \in S$
2)$\forall (a,b) \in S\times S$ is uniquely mapped by (*) to some $c \in S$
Also, we define $a*b = a\times b$ [multiplication on reals]
Note that condition 1 and 2 are satisfied.
Let $H \in S$ Clearly, $\frac{1}{H} \in S$. We apply the multiplication operation on them. Thus we have $H*\frac{1}{H} = 1\in S$. NOTE:This is true for any arbitrary $H \in S$. We now define $H > r \forall r \in S$. Note that this is the formal condition for $H$ tending to infinity. Now suppose $|\frac{1}{H}|$ is finite and non-zero[non-zero because it is a part of $S$] . This implies that $H*\frac{1}{H} \to +\infty$ or $-\infty$ as $H$ grows larger. But this is impossible as long as $H \in S$ [product = 1]. Thus, the only option is to make $|\frac{1}{H}|$ infinitesimally small [it can't be infinitesimally large either]
Hence proved.
