Solution of A Continuous Injective Function on a Closed Interval is Strictly Monotone

Since $a\ne b$ and $f$ is injective, $f(a)\ne f(b)$. The order of $\mathbb{R}$ is total, so either $f(a)<f(b)$ or $f(b)<f(a)$.

Step 1 (key step). We claim: if $p,q\in[a,b]$ satisfy $p<q$ and $f(p)<f(q)$, and if $p<z<q$, then

$$f(p)<f(z)<f(q) .$$

First, $z\in[a,b]$, since $a\le p<z<q\le b$. As $z\ne p$ and $z\ne q$, injectivity gives $f(z)\ne f(p)$ and $f(z)\ne f(q)$.

Suppose $f(z)<f(p)$. Then $f(z)\le f(p)\le f(q)$. The restriction of $f$ to $[z,q]$ is continuous on $[z,q]$ by clause 1 of Restriction Stability of Continuity and of the Derivative, and $z<q$; so Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals, applied on $[z,q]$ with the value $f(p)$, gives $w\in[z,q]$ with $f(w)=f(p)$. But $p<z\le w$, so $w\ne p$, and injectivity is contradicted.

Suppose instead $f(q)<f(z)$. Then $f(p)\le f(q)\le f(z)$. The restriction of $f$ to $[p,z]$ is continuous on $[p,z]$, and $p<z$; so Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals, applied on $[p,z]$ with the value $f(q)$, gives $w\in[p,z]$ with $f(w)=f(q)$. But $w\le z<q$, so $w\ne q$, and injectivity is contradicted.

Since the order is total and $f(z)$ equals neither $f(p)$ nor $f(q)$, the only remaining possibility is $f(p)<f(z)$ and $f(z)<f(q)$, which is the claim.

Step 2: the case $f(a)<f(b)$. Assume $f(a)<f(b)$, and let $x,y\in[a,b]$ with $x<y$. We show $f(x)<f(y)$.

Since $x<y\le b$ we have $x<b$. We first check that $f(x)<f(b)$. If $x=a$, this is the assumption. If $a<x$, then $a<x<b$, and Step 1 with $p=a$, $q=b$, $z=x$ gives $f(a)<f(x)<f(b)$, so in particular $f(x)<f(b)$.

Now $x<b$ and $f(x)<f(b)$. If $y=b$, then $f(x)<f(b)=f(y)$. If $y<b$, then $x<y<b$, and Step 1 with $p=x$, $q=b$, $z=y$ gives $f(x)<f(y)<f(b)$, so in particular $f(x)<f(y)$.

Hence $f(x)<f(y)$ whenever $x<y$, that is, $f$ is strictly increasing on $[a,b]$ in the sense of Monotone Real Function (iii).

Step 3: the case $f(b)<f(a)$. Assume $f(b)<f(a)$, and define $g:[a,b]\to\mathbb{R}$ by $g(x)=(-1)\cdot f(x)$. By clauses 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, $g$ is continuous on $[a,b]$. It is injective: if $x,y\in[a,b]$ with $x\ne y$ then $f(x)\ne f(y)$, hence $-f(x)\ne-f(y)$, that is, $g(x)\ne g(y)$. Multiplying $f(b)<f(a)$ by $-1$ reverses the inequality, by Elementary Order Arithmetic in an Ordered Field, so

$$g(a)=-f(a)<-f(b)=g(b) .$$

Steps 1 and 2 apply verbatim to $g$ in place of $f$ and show that $g$ is strictly increasing on $[a,b]$. Hence, for $x,y\in[a,b]$ with $x<y$, we have $-f(x)<-f(y)$ and therefore $f(y)<f(x)$, again by Elementary Order Arithmetic in an Ordered Field. That is, $f$ is strictly decreasing on $[a,b]$ in the sense of Monotone Real Function (iv).

In both cases $f$ is strictly monotone on $[a,b]$.