11 Theorem Let $f$ and $g$ be isotone functions from a chain $X$ to a chain $Y,$ let $X_0$ be a subset of $X$ on which $f$ and $g$ agree, and let $Y_0$ be $f[X_0]$. A sufficient condition that $f= g$ is that $Y_0$ intersects every set of the form $\{y: u\lt y\lt v, ~u\ne y ~~\textrm{and}~~ y \ne v\},$ where $u$ and $v$ are members of $Y$ such that $u\lt v.$
Proof If $f\ne g,$ then $f(x) \ne g(x)$ for some $x$ in $X,$ and we may suppose that $f(x)\lt g(x)$. Each point of $X_0$ which is less than or equal to $x$ maps under $f$ to a point less than or equal to $f(x)$, because $f$ is isotone, and each point which is greater than or equal to $x$ maps under $g$ into a point greater than or equal to $g(x)$, because $g$ is isotone. It follows that no point of $X_0$ maps into the set $\{y: f(x)\lt y\lt g(x), f(x)\ne y~~\textrm{and}~~y \ne g(x)\}$, and the theorem is proved.
The excerpt is taken from Orderings in General Topology by John Kelley.
I'm not comprehending the bold statements above in the excerpt.
At the first sight, I can conclude that the proof is using the method of contradiction.
What is the point of taking in consideration the function $g$ when the points $\geq x\in X_0\,?$ I'am simply not getting that.
How did Kelley conclude then that $X_0$ doesn't map to $\{y: f(x)\lt y\lt g(x) \}\,?$
I'm failing to visualise that how it proves $Y_0= f[X_0]$ intersects with $\{y: u\lt y\lt v\}.$
About the first two bold statements, I think they already have the explanation there (because $f$ and $g$ are isotone). Perhaps you overlooked something here...
For the final part, let $z \in X_0$ and suppose $$f(x) < f(z) = g(z) < g(x).$$
Then it must be $x < z$, because $X$ is a chain.
But then $g(z) < f(x)$ is in contradiction with the isotonicity of $g$.
So there is no $z \in X_0$ such that $$f(x) < f(z) < g(x),$$ or equivalently, there is no $y$ in the image of $X_0$ by $f$ (or $g$, which is the same) such that $$f(x) < y < g(x)$$ and this is certainly the case if $f(X_0) = g(X_0) = Y_0$ every open interval of the chain.