Prob 10 Sec 26 in Munkres' TOPOLOGY, 2nd ed: How to give examples of this result failing? 
Let $X$ be a compact topological space. Let $f_n \colon X \to \mathbb{R}$ be a sequence of continuous functions such that $f_n(x) \leq f_{n+1}(x)$ for all $x \in X$ and for all $n \in \mathbb{N}$. Let $f \colon X \to \mathbb{R}$ be a continuous function such that, for each $x \in X$, the sequence $f_n(x) \to f(x)$ as $n \to \infty$. Then the sequence $f_n$ converges uniformly to $f$ on $X$. 

This is problem 10 (a), Sec. 26, in the book Topplogy by James R. Munkres, 2nd edition. I've managed to give a proof of this result. 
However, what if we require that $f_n(x) \geq f_{n+1}(x)$ for all $x \in X$ and for all $n \in \mathbb{N}$? My assertion is the convergence is still uniform. 
Here's my proposed proof: 

Let $\epsilon > 0$ be given. Let $x \in X$. Since $f_n(x) \to f(x)$ as $n \to \infty$, there exists $N(x) \in \mathbb{N}$ such that $\vert f_n(x) - f(x) \vert < \epsilon$ if $n \geq N(x)$. 
Since $f_n(x)$ is a monotically decreasing sequence, we must have $f_n(x) \geq f(x)$ for all $n$. So
  $$\vert f_n(x) - f(x) \vert = f_n(x) - f(x) < \epsilon \ \ \ \mbox{ if } \ n \geq N(x). $$
  In particular, 
  $$f_{N(x)} (x) - f(x) < \epsilon.$$
  Since $f_{N(x)} - f \colon X \to \mathbb{R}$ is continuous, the set 
  $$U_x \colon= \left( f_{N(x)} - f \right)^{-1} \left( \ (-\infty, \epsilon) \ \right)$$
  is open in $X$ and the point $x$ belongs to $U_x$. 
In this way, we have an open covering $\{ U_x \colon x \in X \}$ of $X$. Since $X$ is compact, there is a finite subcollection $U_{x_1}, \ldots, U_{x_k}$ that also covers $x$. 
Let $N \colon= \max \left( N(x_1), \ldots, N(x_k) \right)$. 
Let $y \in X$. Then $y \in U_{x_i}$ for some $i = 1, \ldots, k$. So, $$f_{N(x_i)} (y) - f(y) < \epsilon. $$
So, if $n \geq N$, then we have 
  $$
\begin{align*}
\vert f_n(y) - f(y) \vert &= f_n(y) - f(y) \\
&\leq f_N(y) - f(y) \\
&\leq f_{N(x_i)} (y) - f(y) \\ 
&< \epsilon. 
\end{align*}
$$
Thus, for every given $\epsilon > 0$, there exists a natural number $N$ such that, if $n \geq N$, then we have 
  $$\vert f_n(y) - f(y) \vert < \epsilon$$
  for all $y \in X$. 
Hence $f_n$ converges uniformly to $f$ on $X$. 

Is the above proof correct? 
But the result breaks down in the following example. 
Let $X \colon= [0,1]$, and, for each $n \in \mathbb{N}$,  let $f_n \colon X \to \mathbb{R}$ be defined by 
$$f_n(x) \colon= x^n \ \ \ \mbox{ for all } \ x \in X.$$
Now, for all $n \in \mathbb{N}$, 
$$f_n(0) = 0 = f_{n+1}(0),$$
and 
$$f_n(1) = 1 = f_{n+1}(1);$$
for  $x \in (0,1)$, we have 
$$f_n(x) = x^n > x^{n+1} = f_{n+1}(x).$$
The sequence $f_n$ converges pointwise to the function $f \colon X \to \mathbb{R}$ defined by 
$$
f(x) \colon= 
\begin{align*}
0 & \ \mbox{ if } \ 0 \leq x < 1; \\
1 & \ \mbox{ if } \ x = 1. 
\end{align*}
$$
but this convergence is not uniform. But $f$ is not continuous. So the hypotheses of our result are not met fully. 
(i) What if $X$ is not compact?
(ii) What if the neither of the monotonicity conditions is satisfied? 
Please give an elementary enough example in each case where the result breaks down. 
 A: *

*The proof seems ok. A simple proof would be setting $g_n:=-f_n$ and using the result for monotonically increasing sequences. 

*The example of the powers you gave with $X=[0,1]$ is not really a counter-example as the limit function $f$ is not continuous.

*If you use the same power functions with the non-compact domain $X:=[0,1)$ you'll find a sequence of continuous functions which converges to a continuous limit function non-uniformly.

*Define $f_n:[0,1] \to \mathbb{R}$ as the piecewise linear function passing through the points $(0,0),(1/n,1),(2/n,0),(1,0)$. The sequence $(f_n)_{n=2}^\infty$ is a sequence of continuous functions that converge uniformly to the continuous zero function. However, the convergence is not uniform.
A: Your example of $f_n(x) = x^n$ also shows that compactness is necessary, if you restrict the domain to the noncompact set $[0,1)$.  Then $f_n(x)$ converges pointwise to the continuous function $f(x) = 0$.
To show that something like monotonicity is necessary, consider $X = [0,1]$, and set 
$$
f_n(x) = \begin{cases}
0 & x \le 1- \frac{2}{n} \\
nx-n+2 & 1-\frac{2}{n} < x \le 1 - \frac{1}{n}\\
n-nx & 1-\frac{1}{n} < x \le 1\\
\end{cases}
$$  Here the nontrivial part of the graph of $f_n$ is an increasingly narrow triangle of height 1.
For all $x\in [0,1]$, $f_n(x) \to 0$, but since $f_n\left(1-\frac 1 n\right) = 1$, the convergence is not uniform.
A: 1) Counterexample when $X$ is not compact:
Take $X=\mathbb{R}$ with the usual topology, and let $f_n(x) = \min \{ x,n\}$.
Then $f_n \leq f_{n+1}$ for all $n$, and $(f_n)$ converges to the function $f(x)=x$ pointwise, but certainly not uniformly (since $\sup_{x\in \mathbb{R}} |f_n(x)-x| = \infty$ for all $n$).
2) Counterexample when monotonicity is not satisfied:
Let $X = [0,1]$, and define $f_n(x) = \frac{nx}{1+n^2x^2}$.
Then $f_n \to 0$ pointwise on $[0,1]$, so the limit is continuous, but convergence is not uniform since $f_n(\frac{1}{n}) = \frac{1}{2}$, which does not converge to $0$.
