Dini's theorem and monotonicity Is Dini's theorem can be applied for a function series such that for every $x$ there's an $N_x$ where for every $n>N_x:$ $f_n(x) \ge f_{n+1}(x)$; Meaning, the monotonicity start somewhere in the series, for every $x$. 
 A: It is not sufficient to conclude uniform convergence if the sequence only becomes eventually monotonic at each point. That does not prevent a sliding hump like we have with the functions
$$f_n(x) = \begin{cases} \quad nx &, 0 \leqslant x \leqslant \frac{1}{n} \\ 2 - nx &, \frac{1}{n} < x \leqslant \frac{2}{n} \\ \quad 0 &, \frac{2}{n} < x \end{cases}$$
on $[0,1]$, for $n \in \mathbb{N}\setminus \{0\}$. We have $f_n(x) \to 0$ for all $x$, so the limit function is continuous, and $f_n(1/n) = 1$, so the convergence is not uniform. If we set $N_0 = 1$ and $N_x = \lceil 1/x\rceil$ for $x > 0$, then the sequence $\bigl(f_n(x)\bigr)_{n\geqslant N_x}$ is monotonic for all $x$, so the sequence satisfies the conditions.
Of course the convergence is uniform if we demand that there is an $N$ such that the sequence $\bigl(f_n\bigr)_{n \geqslant N}$ is monotonic, but we need a uniform index at which the sequence becomes monotonic, the index can't (arbitrarily) depend on the point.
A: In the version of Dini's theorem that I had, it was already consider this case. I attach you the proof, with your case conisdered. I hope it's useful.
Claim:
Let $(E,d)$ be a compact metric space and $f_n:E \to\mathbb N$ a succession of continuous functions. Let $\{f_n\}_n$ converge pointwise to a continuous function $f$. And, for every $x\in E$ it should exist an $N_x \in \mathbb N$ such that for every $n>N_x$, $f_n(x) \geq f_{n+1}(x)$. 
Then $\{f_n\}_n$ converge uniformly to $f$.
Proof:
Let $\epsilon>0$ and $x \in E$ be given.


*

*$\exists N_x$ such that $\forall n>N_x$
$$f_{n+1}(x) \leq f_n(x) \Rightarrow f(x) \leq \cdots \leq f_{n+2}(x)\leq f_{n+1}(x) \leq f_n(x) \Rightarrow$$
$$\Rightarrow f(x) -f_n(x) \leq 0 \Rightarrow f(x) -f_n(x) < \epsilon/3$$ 

*Because $f$ and $f_n$ are continuous, it exists and open neighbourhood $V(x)$ of $x$ such that if $y \in V(x)$ then 
$$|f(x)-f(y)|<\epsilon/3$$ and 
$$|f_n(x)-f_n(y)|<\epsilon/3$$
On the other hand:
$$ |f(y)-f_n(y)| = |f(y)-f(x)+f(x)-f_n(x)+f_n(x)-f_n(y)| \leq  $$
$$|f(y)-f(x)| + |f(x)-f_n(x)| + |f_n(x)-f_n(y)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$$
Because $\{V(x) : x \in E\}$ is an open cover of the compact $E$, then it exist a finite subcover $\{V(x_1),V(x_2), \cdots, V(x_p)\}$.
If we define $N:=max \{ N_{x_1},N_{x_2},\cdots,N_{x_p} \}$. Then, for each $x \in E$, $x$ will also belong to one $V(x_i)$ and then $\forall n \geq N$
$$|f(x)-f_n(x)|<\epsilon  (\forall x \in E)$$
Therefore $\{f_n\}_n$ converge uniformly to $f$.
