Proof that there are infinitely many positive rational numbers smaller than any given positive rational number.

Your proof does not work. Indeed, subtracting $1$ from $\frac p q$ will give you a rational number, but it will be negative by assumption, so this doesn't help you (since it doesn't give you a contradiction).

A simpler approach: Explicitly state what the infinitely-many positive rationals less than $x$ are.

Hint: If $y$ is a positive rational, what can you say about $\frac{y}2$? About $\frac{y}4$? $\frac{y}8$? ...

Before diving into a proof right away, it might help to think of some examples and generalizing a pattern you find. For example, suppose $x=1/3$. Can you think of a smaller positive rational number? Well, we could do a bunch of things to make a fraction smaller. We could try decreasing the numerator $-$ but as you discovered, things might not work out if the numerator is already very small. Instead, we can try increasing the denominator. Indeed, since the integers are unbounded above, this is always possible.

Proof. We proceed by contradiction. Suppose instead that there are only finitely many positive rational numbers less than $x$. Now let $y$ be the smallest of these positive rational numbers. Observe that $y=p/q$ where $p,q\in \mathbb{Z}$ and $p,q>0$. Now consider the number: $$ z=\dfrac{p}{q+1} $$ Observe that since $q+1\in \mathbb{Z}$ and $q+1>1>0$, we know that $z$ is a positive rational number. But then we have $z<y$, contradicting the minimality of $y$. Hence, there are infinitely many positive rational numbers less than $x$, as desired.

First, you don't need Case II. If $x\in\mathbb Q_{>0}$, then you can assume, that $x=\frac{p}{q}$, where $p,q>0$.

Your general idea is good: Assume, that $x$ is smallest possible and find an even smaller one, which then is a contradiction. Now let's answer your questions:

2) You already noticed, that you only proved that there is a smaller rational number, not necissarily positive. Your proof is basically "If $x$ is the smallest, then $x-1$ is smaller, a contradiction."

1) Of course, this a valid operation, it actually disproved your assumption, that $x$ was the smallest rational number.

3) is right, too. With an arbitrary $k$, you get infintely many smaller rationals.

To prove the positive case, notice, that if $x\in\mathbb Q_{>0}$, then $0<\frac{x}{k}<x$ for all $k\in\mathbb Z$.

There is no way to justify the subtraction of $1$: If $\frac{p}{q}\lt 1$, and $p$ and $q$ are positive, then indeed you showed correctly that $\frac{p-q}{q}\lt \frac{p}{q}$.

However, you did not show that $\frac{p-q}{q}$ is positive, and it will not be, for instance, if $p=3$ and $q=5$.