In the Dini's Theorem, On the compact set $K$,if $f_n$ is a sequence of monotone increasing or decreasing continuous functions, i,e $f_n(x)\leq(\geq) f_{n+1}(x)$ for all $n$ and all $x$, converges pointwisely to a continuous function $f$, then the convergence is uniform.
But now if we only know that each continuous function $f_n$ is monotone w.r.t x, but do not know for each $x$, whether $f_n(x)\leq f_{n+1}(x)$, or $f_n(x)\geq f_{n+1}(x)$, then can we prove the convergence is uniform? Or can we find a countereample?
Informal "proof". Suppose $f$ and $f_n$ are increasing. Draw their graphs and turn the paper $45$ degrees clockwise. Now you see a sequence of $1$-Lipschitz (hence equicontinuous) functions converging pointwise. The Arzelà–Ascoli theorem implies that the convergence is uniform.
Now let's be serious. Using the continuity of $f$, partition its domain into subintervals on each of which the oscillation of $f$ is less than $\epsilon$. For large $n$, the difference $|f_n-f|$ is less than $\epsilon$ at every partition point. By monotonicity, $|f_n-f|\le 2\epsilon$ everywhere.
Explanation of the latter: say, the values of $f$ on some subinterval are between $a$ and $a+\epsilon$. Since $|f_n-f|\le \epsilon$ at the endpoints, and $f_n$ is monotone, we have $a- \epsilon \le f_n \le a+2\epsilon$ on the subinterval. Conclusion follows.