On the existence of infinitesimals. Creating the first infinitesimal by use of Compactness Theorem
In this question someone tried to create the first infinitesimal using the compactness theorem, but I don't know much about compactness theorem and model theory. But the second answer in that question says that no need to learn model theory or compactness theorem to prove the existence of infinitesimals. He said something about ultrafilters and taking ultraproduct on real numbers and I don't even know what that is, but I am highly interested to know about it so please explain me what exactly he says in that answer or provide any related pdf's or videos to learn it.
 A: Consider the polynomials with variable (or "indeterminate") $x$, with real (or just rational) coefficients. We can give them an ordering by saying that a polynomial is "positive" if its term of lowest degree has a positive coefficient. In analogy with "leading coefficient", I'll refer to this as the "trailing coefficient"*. So $5$, $-x+4$, and $-2x^7+3x$ are all positive because of their trailing coefficients $5,4,3$. But $6x^2-4x$ is negative because its trailing coefficient is $-4$, etc.
To compare two polynomials $f(x)$ and $g(x)$, just take the difference $f(x)-g(x)$ and see if it's the zero polynomial (so $f(x)=g(x)$), has positive trailing coefficient (so $f(x)>g(x)$), or has negative trailing coefficient so that $g(x)-f(x)$ would be positive (so $f(x)<g(x)$).
With this ordering, note that $0<x$ since $x-0=1x$ is positive. But we also have $x<\dfrac19$ because $\dfrac19-x=-x+\dfrac19$ has positive trailing coefficient. This would work with any positive real number in place of $\dfrac19$, so in our ordered polynomials, $x$ is "infinitesimal": positive, but smaller than $\dfrac1n$ for any positive integer $n$. $x^2$ is smaller than $rx$ for any positive real $r$, etc.
The polynomials with this order act as a sort of "number system" in that you can add, subtract, and multiply. And those operations "play nicely" with the order: for instance, the sum or product of two positive polynomials is positive. (The technical term for this structure is ordered ring. And since it has infinitesimals, it's non-Archimedean.)

You may also want division in your number system, in which case you can form fractions of polynomials (sometimes called "rational functions"). An expression of the form $\dfrac{f(x)}{g(x)}$ is allowed if $g(x)$ is not the zero polynomial. Then we can compare these fractions just as in arithmetic. To compare $\dfrac{f(x)}{g(x)}$ and $\dfrac{p(x)}{q(x)}$, consider the polynomial $h(x)=q(x)f(x)-p(x)g(x)$: If $h(x)$ is zero, then $\dfrac{f(x)}{g(x)}=\dfrac{p(x)}{q(x)}$; if $h(x)>0$ then $\dfrac{f(x)}{g(x)}>\dfrac{p(x)}{q(x)}$, and analogously for $h(x)<0$. If we add and multiply fractions as usual, the arithmetic operations and this order are again compatible. (The technical term for this construction is the field of fractions, and so we now have a non-Archimedean ordered field.)
With this setup, we find that $\dfrac{1}x>9$ since $1*1-9*x=-9x+1$ has positive trailing coefficient. This works if we replace $9$ with any other real number, so $\dfrac1x$ is "infinite" in that it is greater than any integer. Similarly, $\dfrac{-1}{x}$ is less than any real number. $\dfrac1{x^2}$ is larger than $\dfrac{r}{x}$ for any real $r$, etc.

(This answer was inspired by the answer of egreg to Ordering the field of real rational functions. If comparing things, note that the $x$ in that question is infinite, acting like the reciprocal of my $x$.)
*The choice of "trailing coefficient" aligns with the answer of 5xum to How to name the opposite of "leading term" in a polynomial?.
