How to approach this proof problem, what proof to use, what assumption to use? This is a problem from Discrete Mathematics and its Applications

Here is the definition of rational that my book uses 

Usually when I approach this type of a problem, I can find a type of proof to use and what assumption to make. But for this one I can't find one that works
First I tired a direct proof. That is, I assumed $r$ is irrational or there do not exists integers $p$ and $q$ with $q\neq 0$ such that $r = \frac{p}{q}$. But I couldn't go from that to saying $|r - n| < \frac{1}{2}$ where $n$ is some integer.
I also tried proof by contradiction. The initial problem is proving $p \to q$ so proof by contradiction would be showing that $\neq (p\to q)$ or $p\land \neg q$ leads to a contradiction. I didn't know how to get from $r$ being a rational number to there not being an integer $n$ for which $|r - n| < 1/2$.
Lastly, I tried proof by contraposition, which would be $\neg q \to \neg p$. If there is not a unique integer $n$ such that the distance between $r$ and $n$ is less than $1/2, r$ is rational. But how would you get from for all integers $|r-n| \geq 1/2$ to $r$ is rational?
 A: HINT: If $x$ is irrational, then $x$ is not an integer, so there is a unique integer $n$ that $n<x<n+1$. Clearly the distance from $x$ to the closer of $n$ and $n+1$ is at most $\frac12$. If $x$ is irrational, can that distance be $\frac12$?
Added: Please take note also of Jack M’s excellent comment under the original question; I was a bit rushed when I wrote my answer, or I’d have tried to make the same point. Picking a proof technique first is usually a bad idea: the nature of the problem itself and any intuitions that you can form about it should guide the choice of proof technique to try.
A: For every real number $r$ except members of $\{\pm0.5, \pm1.5,\pm2.5,\pm3.5,\ldots\}$ there is a unique integer $n$ such that the distance between $r$ and $n$ is less than $1/2$.
If $r$ is irrational, then $r\not\in\{\pm0.5, \pm1.5,\pm2.5,\pm3.5,\ldots\}$.
A: Given $r$, let $a$ be the closest integer to $r$ less than $r$, and let $b$ be the closest integer to $r$ greater than $r$. In the notation introduced in Section 1.8 [I also have this book], we have $a=\lfloor r \rfloor$ and $b=\lceil r \rceil$. In fact, $b=a+1$. The distance between $r$ and any integer other than $a$ or $b$ is greater than $1$ so cannot be less than $\frac{1}{2}$. Furthermore, since $r$ is irrational, it cannot be exactly half-way between $a$ and $b$, so exactly one of $r-a < \frac{1}{2}$ and $b-r < \frac{1}{2}$ holds.
