Trouble understanding how the Transfer Principle is applied for the Extreme Value theorem.

For me the problem here is that the statement Keisler gives of the transfer principle doesn't quite fit with how it is being used. You might want to look at a more formal source to clarify exactly what transfer says (try Goldblatt, or https://en.wikipedia.org/wiki/Transfer_principle and the references it cites).

Here transfer is being applied to the statement "if $n$ is a natural number and $s_1,\ldots, s_n$ are reals then $\{s_1,\ldots,s_n\}$ has a maximum." Transfer (in its full form - not necessarily in the way Keisler states it) tells you that this applies for $n \in \mathbb{N}^*$ too, which is exactly what is needed in the proof.

This can seem confusing if you think of nonstandard natural numbers as "infinitely large," because it is certainly not true that an infinite subset of $\mathbb{R}^*$ has to be bounded. This application of transfer tells us only that if $\nu$ is any natural number, even a nonstandard one, then every sequence $s_1,\ldots, s_\nu$ is bounded.

It is helpful to work through an example of a discontinuous unbounded function on a compact interval, to see why Keisler's argument wouldn't apply to that. Let's take $f(0)=0$ and $f(x)=1/x$ for $x>0$, so that $f: [0,1] \to \mathbb{R}$ is unbounded. We start by picking a partition of $[0,1]$ with evenly spaced points $0,1/H,2/H,\ldots, (H-1)/H, 1$, where $H$ is an "infinitely large" natural number. There is indeed a partition point at which $f$ is maximal, namely $f(1/H)=H$. The standard part of $1/H$, which gets called $c$ in Keisler, is 0. But you can't get any relationship betweeen $f(c)$ and $f(1/H)$, even though $c$ and $1/H$ are infinitely close, because $f$ is not continuous at 0. This means the argument breaks down, as it must.

Let me propose a different interpretation of transfer (in the formulation suggested by m_t_) for the Intermediate Value Theorem.

Consider the following "standard" argument about ordinary real numbers. For all $n \in \mathbb{N}$, it is possible to partition $[a,b]$ into $a, a+\frac{b-a}{n}, \ldots, a+n\frac{b-a}{n}=b$. Since there is a finite number of partition points (they are $n+1$), it is well-defined the maximum among the values $f(a), f\left(a+\frac{b-a}{n}\right), \ldots, f(b)$.

Recall that, by transfer, finite and $^\ast$finite sets satisfy the same properties. As a consequence, you can always pick the biggest element of a $^\ast$finite set. Hence, transfer entails also that for all $n \in\ \! ^\ast\mathbb{N}$ the number $$\max\left\{ f(a), f\left(a+\frac{b-a}{n}\right) , \ldots, f\left(a+n\frac{b-a}{n}\right)=f(b) \right\}$$ is well-defined for every partition of $^\ast[a,b]$ into $n$ equal parts. In other words, you are proving the Extreme Value Theorem by applying transfer to the statement "every finite set has a maximum element", and not to the Extreme Value Theorem for real numbers.

Let now $n \in\ \! ^\ast\mathbb{N}$ be infinite, and let $$f\left(a+K\frac{b-a}{n}\right) = \max\left\{ f(a), f\left(a+\frac{b-a}{n}\right) , \ldots, f\left(a+n\frac{b-a}{n}\right)=f(b) \right\}.$$ (What I call $\frac{b-a}{n}$ is the infinitesimal $\delta$ in the original proof by Keisler). Define $c=\ \!^\circ\left(a+K\frac{b-a}{n}\right)$. By continuity of $f$ you have that $^\circ\left(^\ast f\left(a+K\frac{b-a}{n}\right) \right) = f(c)$. From here onward, it is possible to follow the original proof by Keisler.