$\Phi$ is the set of functions $f: [a, b] \to \mathbb{R}$ where $f$ is bounded, non-decreasing and continuous on $[a, b]$. Is $\Phi$ compact with $L^2$ norm?
What about $\Psi$ is the set of functions $f: [a, b] \to \mathbb{R}$ where $f$ is non-decreasing and continuous on $[a, b]$. Is $\Psi$ compact?
No. Consider
$$f_n(x) = \begin{cases}\qquad 0 &, x < \frac{1}{2} - \frac{1}{2^{n+1}} \\ 2^n\left(x-\frac{1}{2}+\frac{1}{2^{n+1}}\right) &, \frac{1}{2}-\frac{1}{2^{n+1}} \leqslant x \leqslant \frac{1}{2}+\frac{1}{2^{n+1}}\\ \qquad 1 &, x > \frac{1}{2} + \frac{1}{2^{n+1}}. \end{cases}$$
The $L^2$-limit of that sequence is
$$f(x) = \begin{cases}0 &, x < \frac{1}{2}\\ 1 &, x \geqslant \frac{1}{2} \end{cases}$$
(Source above counterexample: Limit of continuous sequence of functions with respect to $L^{2}$ norm is continuous.). Then $f_n$ is bounded, continuous and non-decreasing, yet its limit is not continuous. Hence, $\Phi$ is not $L_2$-closed and thus not compact.