According to Dini's theorem:
If $X$ is a compact topological space, and $\{ f_n \}$ is a monotonically increasing sequence (meaning $f_n(x) \leq f_{n+1}(x)$ for all $n$ and $x$) of continuous real-valued functions on $X$ which converges pointwise to a continuous function $f$, then the convergence is uniform.
The same conclusion holds if $\{ f_n \}$ is monotonically decreasing instead of increasing.
(Note: I have proven both cases)
1) If for every $n$ $\{f_n(x0)\}$ is monotonic but for some values of $n$ it's monotonically decreasing and for other it's monotonically decreasing. for example; for all even values it is increasing and for non-even values it is decreasing.
How could I prove that Dini's theorem is effective in this case? In other words, how to prove that the convergence is uniform
2) Can I require that {fn} is monotonic starting from a specific index that's related to x? for example; what if {fn} is monotonic when n>7x?
Note: I already got an answer to no. 2 thanks to @Robert
(1) Do you mean a sequence $f_n$ such that for each $n$, either $f_{n+1} \ge f_n$ everywhere or $f_{n+1} \le f_n$ everywhere?
Let $g_n$ be a sequence of functions on $X$ with $g_n > 0$ and $g_n \to 0$ pointwise but not uniformly on $X$, take $f_{2n}(x) = g_n(x)$ and $f_{2n+1}(x) = -g_n(x)$. Then $f_{n+1} < f_n$ if $n$ is even and $f_{n+1} > f_n$ if $n$ is odd, and $f_n \to 0$ pointwise but not uniformly.
EDIT: OK, it seems that wasn't what you meant. If $f_n$ is monotonically decreasing on $A \subset X$ and monotonically increasing on $B \subset X$, with $A \cup B = X$, note that we can take $A$ and $B$ to be compact (since $\{x \in X: f_{n+1}(x) \le f_n(x)\}$ and $\{x \in X: f_{n+1}(x) \ge f_n(x)\}$ are closed). Thus $f_n$ converges uniformly to $f$ on $A$ and converges uniformly to $f$ on $B$, and therefore converges uniformly to $f$ on $A \cup B = X$.
(2) Consider a sequence of triangular bump functions on $[0,1]$
$$ f_n(x) = \cases{n x & if $0 \le x \le 1/n$ \cr 2 - nx & if $1/n \le x \le 2/n$ \cr 0 & if $x \ge 2/n$\cr} $$ Thus for any $x > 0$, $f_{n+1}(x) \le f_n(x)$ when $n \ge 1/x$, while $f_{n+1}(0) \le f_n(0)$ for all $n$. $f_n \to 0$ pointwise but not uniformly.