Setup
Background
I like Terry Tao's voting analogy for (Robinson's) hyperreals. Basically, a hyperreal is a sequence of reals that vote each time you ask about a property (like "are you bigger than 5?"). And how to determine which infinite collections of voters count as good majorities is handled by some technical "ultrafilter" stuff, only some of whose properties are needed to answer your question.
Properties
(Leaning on this analogy, I will use "is a majority" to mean "is an element of the relevant ultrafilter".)
Property 1: If $S$ is a subset of the naturals, either $S$ or its complement $\mathbb{N}\setminus S$ is a majority.
Property 2: If $S$ and $T$ are majorities, then their intersection $S\cap T$ is a majority.
Property 3: The empty set $\varnothing$ is not a majority.
Do we have $A\gtreqqless B$?
Will one of the three statements A >/=/< B always be true?
Yes, we can prove this with the three properties mentioned above. Suppose $A=\left[(a_1,a_2,\ldots)\right]$ and $B=\left[(b_1,b_2,\ldots)\right]$.
"Dichotomy"
First, we will prove that exactly one of $A\le B$ and $A>B$ holds.
At least one
We either have 1. $a_i\le b_i$ in a majority of cases ( in which case $A\le B$ ). Or we have 2. $a_i\le b_i$ in a non-majority of cases so that we have $a_i>b_i$ in the complement cases, which must be a majority by Property 1 ( so that $A>B$ ).
At most one
We should check whether we could have both $A\le B$ and $A>B$ at the same time. Suppose we did: Then we would have both $a_i\le b_i$ and $a_i>b_i$ each happening in a majority of cases. Then we could use Property 2 to intersect those majorities to learn that "$a_i\le b_i$ and $a_i>b_i$" (a contradiction) happens in a majority of cases. But Property 3 says the empty set can't be a majority, so "$A\le B$ and $A>B$" is impossible.
In conclusion, exactly one of "$A\le B$" or "$A>B$" must hold.
Trichotomy
Note that since $A$ and $B$ were arbitrary in the above argument, we can swap $A$ and $B$ to also learn that exactly one of "$B\le A$" or "$B>A$" must hold.
Now we can show that exactly one of "$A<B$", "$A=B$", or "$A>B$" must hold.
At most one
Suppose we had $A>B$ and either of $A<B$ or $A=B$. Then we'd have $A>B$ and $A\le B$, which we already know to be impossible.
Now suppose we had $A<B$ and $A=B$. Then we'd have $B>A$ and $B\le A$, which we also already know to be impossible.
At least one
We can break things up into cases based on which pairs of statements above are true.
Case I: $A\le B$ and $B\le A$
In this case, a majority of the time we have $a_i\le b_i$ and a majority of the time we have $b_i\le a_i$. By Property 2, the intersection is still a majority, so we have "$a_i\le b_i$ and $b_i\le a_i$" a majority of the time. In other words, $a_i=b_i$ a majority of the time, so $A=B$.
Case II: $A\le B$ and $B>A$
In this case, we know $B>A$, which we may wish to write as $A<B$.
Case III: $A>B$ and $B\le A$
In this case, we know $A>B$.
Case IV: $A>B$ and $B>A$
In this case as well, we know $A>B$. (Though we also know this case to be impossible.)
Complicated comparisons
what about more complicated sequences like the primes {2, 3, 5, 7, 11...}
A lot of deep mathematics can be phrased in terms of comparisons of sequences in this way.
For example, suppose that $P=\left[(2,3,5,7,11,\ldots)\right]$ and $T=\left[(2,4,8,16,32,\ldots)\right]$. If all you know is that $p_n<4^n$ (as in Elementary bounds on the primes), then you won't be able to tell whether $P<T$ or $P>T$ is true (though you'd know that $P=T$ isn't true since only the first prime is a power of $2$). But if you know some more advanced results (see Bounds for $n$-th prime), then you'd know that $P<T$ is true.
For one more example, suppose that $Z$ is defined as the sequence of real parts of non-trivial zeros of the Riemann Zeta function. If you could prove that "$Z=\left[\left(\frac12,\frac12,\frac12,\ldots\right)\right]$ (no matter what ultrafilter is used)", that would be very near to the famous Riemann hypothesis, which carries a million dollar prize.
"Simple" comparisons
what about ... the periodic sequence {2, 1, 2, 1, 2...} ?
By Property 1, either the even indices or the odd indices form a majority. So either $\left[(2,1,2,1,2,\ldots)\right]=\left[(2,2,2,\ldots)\right]$ or $\left[(2,1,2,1,2,\ldots)\right]=\left[(1,1,1,\ldots)\right]$ must hold. However, there are many suitable ultrafilters/designations of "majority". In some the even indices will form a majority. And in others, the odd indices will.
Typically, we don't try to specify which will, since the useful properties of the hyperreals don't depend on this sort of detail. However, it's probably the case that you could pick which of evens/odds you wanted to be a majority. But you won't be able to pick the winner of every "tie" like this (where both one set and its complement are infinite) by hand, so it's probably not worth picking any.
