Integral becomes improper after a substitution

Indeed $d\phi ={dx\over\sqrt{1-x^2}}$, however you do not have a $d\phi$ in your integral, you have $dx=\cos\phi\,d\phi$, so what you are looking at is

$$\int_{\arcsin(\pi/4)}^{\arcsin(\pi/2)}{\cos x\over \sin(\sin(x))}\,dx$$

which is perfectly proper, at least at a glance. But note that what you really did wrong was you ($1$) did the substitution wrong and ($2$) picked a bad function: $\arcsin$ is only defined on $[-1,1]$ and you have used it all the way up to $\pi/2>1$, so it's not really even a valid choice!

To answer the original question, which is psychological more than mathematical, it is perfectly reasonable to intuit that a proper integral should remain proper under substitution. After all, the area does not change.

The source of the impropriety comes from $\sin(x)$ having zero derivative at $x = \frac{\pi}{2}$. Because its value changes very slowly in the vicinity of $\frac{\pi}{2}$, there is a lot of "area" compressed into an infinitesimal change in $\sin(x)$. This is not a concern when you are integrating w.r.t. $x$, but it is when you are integrating w.r.t. $\sin(x)$. To compensate, the value of the integrand must tend to infinity at this point, which results in an improper integral.

Or, consider that $dx = \frac{dx}{dz}dz$. As $\frac{dz}{dx} \rightarrow 0$, $\frac{dx}{dz} \rightarrow \infty $.

Or, consider the slightly simpler case of $$\int_0^1 \left(1-x^2\right) dx$$

Doing a similar substitution $ y := x^2 \implies \frac{dy}{2\sqrt y} = dx$, you get another improper integral:

$$\int_0^1 \frac{1-y}{2\sqrt{y}}dy$$

...and it is again because $\frac{d}{dx}x^2 = 0$ at $x = 0$.

Also, in "$\phi(x):=\arcsin(x)$", asd obviously meant for $\phi$ to be $x$, and $x$ to be the new variable, which is why he later called it $z$ for clarity. The intended meaning was a perfectly valid substitution, as Mathematica confirmed ;)

Your comment under Adam Hughes' answer says you've got $$ \int_{\sqrt{2}^{-1}}^1 {\cos\phi\,d\phi \over \sin(\sin\phi)}. $$ There's nothing improper about that.