 Solution

Set $e_n(x)=|f_n(x)-f(x)|$. The functions $e_n$ are continuous, converge pointwise to zero, and decrease with $n$. This remains true even if the direction of monotonicity of $f_n(x)$ differs between points: at each fixed point a monotone convergent real sequence approaches its limit from one side.

Given $\epsilon >0$, the sets $U_n=\{x:e_n(x)<\epsilon \}$ are relatively open in $[a,b]$, increase with $n$, and cover the interval. Compactness gives a finite subcover; choosing the largest index in it gives $U_N=[a,b]$. For every $n\ge N$ and every $x$, $e_n(x)\le e_N(x)<\epsilon$. Hence $f_n\to f$ uniformly. This proves the required version of Dini's theorem, rather than assuming a common direction of monotonicity everywhere.

On $[0,1)$, take $f_n(x)=x^n$. Each function and its pointwise limit $f=0$ are continuous, and the sequence decreases at every point. NeverthelessThus uniform convergence fails on the noncompact interval although all the other conditions hold.