The following is the statement of Theorem 7.13 on page 150 of Rudin's PMA.
Theorem 7.13 Suppose $K$ is compact and
(a) {$f_n$} is a sequence of continuous functions on $K$,
(b) {$f_n$} converges pointwise to a continuous function on $K$,
(c) $f_n(x)≥f_{n+1}(x)$ for all $x∈K$, $n=1,2,3,…\ $ Then $f_n→f$ uniformly on $K$.
Below the proof, he added a note as follows.
Let us note that compactness is really needed here. For instance, if
$$f_n(x)=\frac{1}{nx+1}\ \ \ \ (0<x<1; n=1,2,3,\ \dots\ )$$ then $f_n(x)\to 0$ monotonically in $(0, 1)$, but the convergence is not uniform.
In his example, that $(0,1)$ is not compact is clear. But I wonder $(i)$ how we can say $f_n(x)\to 0$ monotonically in $(0, 1)$ and (ii) how we can say the compactness is really needed for the convergence of $\{f_n\}$ to be uniform in the given example.
Any participation is appreciated. Thanks.
Edit
After reviewing my question, It is revealed that
(a) Since $\lim_{x\to t}f_n(x)=f_n(t)$ for $x,t\in K=(0,1), n=1,2,3,\ \dots\ $ it follows that {$f_n$} is continuous on $K$
(b) Since $f_n(x)\to 0$ as $n\to \infty$ in $K$, $\{f_n\}$ converges pointwise to a continuous function $f\equiv 0$ on $K$,
(c) Since $\frac{1}{nx+1} \ge \frac{1}{(n+1)x+1}$ for all $x∈K$, $n=1,2,3,…\ $, it follows that $f_n(x)≥f_{n+1}(x)$ for all $x∈K$, $n=1,2,3,…\ $
Thus, we can say that all the three hypotheses, except the compactness, in Theorem 7.13 hold.
Combining (b) and (c), we can say that $f_n(x)\to 0$ monotonically in $(0, 1)$. However
$\left|f_n(x)\right|=\left|\frac{1}{nx+1}\right|\ge \frac{1}{2}$ whenever $nx\le 1$, i.e., $x\le 1/n$ implying that the convergence is not uniform.
Thus we can say that the compactness is needed for the convergence in this example to be uniform.
Am I right?
Hint: Consider the humble graph https://www.desmos.com/calculator/0hmvgkoz2k . The red dot moves the vertical red line sideways and the blue dot shrinks or expands the blue strip vertically. On the control panel on the left, the larger $N$ is the more terms of the sequence are drawn, and the $N$th term is drawn thicker than the previous terms.
The red line depicts pointwise convergence: no matter where it's put, the intersections of it with the graphs of the terms of the sequence are always ordered (as can be seen by clicking on the red line).
The blue strip depicts uniform convergence: in order for the convergence to be uniform no matter how thin the blue strip is there should be a large enough $N$ so that after it all graphs are completely inside the blue strip.
As can be seen, due to noncompactness on the left handside the functions can get stretchy, which allows them to escape the blue strip.
I hope this much is enough to get a sense and write rigorous statements if needed.
Let me note that "compactness is needed" is meant in the sense that it guarantees uniform convergence; there may be a non-compact space $L$ and a sequence of functions on it that satisfy the conditions with a uniform limit (e.g. consider a constant sequence).