Proof Explanation Spivak Calculus: If $f$ is continuous and one-one on an interval, then $f$ is increasing or decreasing on that interval Theorem 2 from Chapter 12 of Spivak's Calculus reads as follows:

If $f$ is continuous and one-one on an interval, then $f$ is increasing or decreasing on that interval,

where one-one means injective and increasing (decreasing) means strictly increasing (decreasing).
The author-provided proof is broken up into three steps: (1), (2), and (3). I do not understand (3) and, in particular, I am not sure I see why it is necessary to continue beyond (1). To me, from (1) we can immediately deduce our desired conclusion.

Spivak's Proof
(1) If $a \lt b \lt c$ are three points in the interval, then either $f(a) \lt f(b) \lt f(c)$ or $f(a) \gt f(b) \gt f(c)$. Suppose, for example, that $f(a) \lt f(c)$. If we had $f(b) \lt f(a)$, then the Intermediate Value Theorem applied to the interval $[b,c]$ would give an $x$ with $b \lt x \lt c$ and $f(x)=f(a)$, contradicting the fact that $f$ is one-one on$[a,c]$. Similarly, $f(b) \gt f(c)$ would lead to a contradiction, so $f(a) \lt f(b) \lt f(c)$. Naturally, the same sort of argument works for the case $f(a) \gt f(c)$.
(2) If $a \lt b \lt c \lt d$ are four points in the interval, then either $f(a) \lt f(b) \lt f(c) \lt f(d)$ or $f(a) \gt f(b) \gt f(c) \gt f(d)$. For we can apply (1) to $a \lt b \lt c$ and then to $b \lt c \lt d$.
(3) Take any $a \lt b$ in the interval, and suppose that $f(a) \lt f(b)$. Then $f$ is increasing: For if $c$ and $d$ are any two points, we can apply (2) to the collection$\{a,b,c,d\}$ (after arranging in increasing order).

Comments
Firstly, it seems to me that we could finish at (1) because (1) is equivalent to demonstrating that $\forall a,b,c \in I\Big [a \lt b \lt c \rightarrow \big[f(a) \lt f(b) \lt f(c) \text{ or } f(a) \gt f(b) \gt f(c) \big] \Big]$, which implies:
$$\forall a,b \in I: a \lt b \rightarrow \left [f(a) \lt f(b) \lor f(a) \gt f(b) \right ]$$
But this is what Spivak means when he says increasing or decreasing. So I am not sure why the proof does not end here.
Looking at (3), was Spivak's objective to show that for any sub-interval $[a,b] \in I$, WLG, if $f(a) \lt f(b)$, for any $c \lt d \in [a,b]$, we must have $f(c) \lt f(d)$? ...meaning that $f$ is (strictly) increasing on $[a,b]$. At which point, because $a$ and $b$ were arbitrary, we can generalize this subinterval?
Any confirmation would be appreciated.
 A: Letting $I$ be the interval that $f$ is continuous and one-to-one (injective) on, from (2) we are provided with:

$\forall a,b,c,d \in I: a \lt b \lt c \lt d \rightarrow [f(a) \lt f(b) \lt f(c) \lt f(d) \text{ or } f(a) \gt f(b) \gt f(c) \gt f(d)]$


Suppose, first, that $I$ is a closed interval defined as $[I_L, I_R]$.
Because $f$ is injective on $I$, we know that $f(I_L) \neq f(I_R)$. So either $f(I_L) \lt f(I_R)$ or $f(I_L) \gt f(I_R)$. WLG suppose the former. Then by (2), letting $a:=I_L$ and $b:=I_R$, we have:

$\forall b, c \in (I_L, I_R): I_L \lt b \lt c \lt I_R \rightarrow f(I_L) \lt f(b) \lt f(c) \lt f(I_R)$

But this implies that:

$\forall b \in (I_L,I_R]: f(I_L) \lt f(b)$ and $\forall c \in [I_L,I_R): f(c) \lt f(I_R)$.

Together, we must have that $\forall a,b \in [I_L,I_R]: a \lt b \rightarrow f(a) \lt f(b)$, which means that $f$ is strictly increasing.

Next, suppose $I$ is an open interval defined as $(I_L,I_R)$. Consider any arbitrary closed subinterval $[a,d]$. WLG, let $f(a) \lt f(d)$. We can apply our above argument to conclude:

$\forall b,c \in [a,d]: b \lt c \rightarrow f(b) \lt f(c) \quad \color{red}{(*_1)}$

Now, consider the values $\alpha_1, \alpha_2$ and $\beta_1,\beta_2$ in $(I_L,I_R)$, where $\alpha_1 \lt \alpha_2 \lt a$ and $d \lt \beta_1\lt \beta_2$.
Because $\alpha_1 \lt \alpha_2 \lt a \lt d$ and $f(a) \lt f(d)$, we can apply (2), which yields $f(\alpha_1) \lt f(\alpha_2) \lt f(a) \lt f(d) \quad \color{red}{(*_2)}$. Similarly, we must have: $f(a) \lt f(d) \lt f(\beta_1) \lt f(\beta_2)\quad \color{red}{(*_3)}$.
Collectively, then, we have that:

$\forall \alpha_1,\alpha_2 \in (I_L,a): \alpha_1 \lt \alpha_2 \rightarrow f(\alpha_1) \lt f(\alpha_2)$ and $\forall \beta_1,\beta_2 \in (d,I_R): \beta_1 \lt \beta_2 \rightarrow f(\beta_1) \lt f(\beta_2)$

Combining this with $\color{red}{(*_{1,2,3})}$, we have:

$\forall a,b \in (I_L,I_R): a \lt b \rightarrow f(a) \lt f(b)$, which means that $f$ is strictly increasing.

