Decreasing function and fixed point

I'm reading a document written about fixed point and decreasing function. In the document, they're considering a lemma that seems to be not right to me. Even the proof they give does not satisfy me.

Lemma: If a function $f$ is strictly decreasing in $K \subset \mathbb{R}$, then $f$ has no more than $1$ fixed point in $K$

Can someone clarify this for me: Why this lemma is true and what's wrong if we change the condition $f$ into strictly increasing?

Edit: I forgot to include the proof from the document. But ye below has clarified my question!!

Solution 1:

If $f$ is strictly increasing, you can have $f(x)=x$, which has infinitely many fixed points.

But if $f$ is strictly decreasing and $f(x_0)=x_0$ for some $x_0\in K$ then, if $x\ne x_0$, then either $x>x_0$ or $x<x_0$. If $x>x_0$, then since $f$ is strictly decreasing, $f(x)<f(x_0)=x_0<x$ and a similar argument shows that if $x<x_0$ we also don't have $f(x)=x$.

Solution 2:

If $a$ is a fixed point of $f$, then $f(a)=a$. If $b$ is another point, if$f$ is strictly decreasing, if $a<b$ then $f(b)<f(a)=a<b$ cannot be fixed, if $a>b$ then $f(b)>f(a)=a>b$ cannot be fixed.