Analysis, Example Function The function needs to fit the following requirements:
Function $f:\mathbb{Q} \to \mathbb{Q}$ where $\vert{f(x)-f(y)}\vert$ $\le$ $1/2$$\vert x-y \vert$ $\forall x,y \in \mathbb{Q}$ and $f(x')\neq x'$ $\forall x'\in \mathbb{Q}$
 A: OP already found an answer, but I will formalize the reasoning here.
Let $f\colon\Bbb Q\to\Bbb Q$ such that $$|f(x)-f(y)|\leq \frac 12|x-y|,\ \forall x,y\in\Bbb Q.\tag{1}$$
Since $\Bbb R$ is metric space completion of $\Bbb Q$ and $f$ is uniformly continuous, there is unique uniformly continuous function $\bar f\colon \Bbb R\to\Bbb R$ such that $\bar f(x) = f(x),\ \forall x\in\Bbb Q$. From continuity it directly follows that for any $x\in\Bbb R$ and any sequence $(x_n)$ in $\Bbb Q$ converging to $x$ in $\Bbb R$, we have $$\bar f(x) = \lim_nf(x_n)$$ and thus the relation $(1)$ is true for $\bar f$ on whole $\Bbb R$ as well. Hence, it is not loss of generality to assume that $f\colon\Bbb R\to\Bbb R$ in the first place, with $f(\Bbb Q)\subseteq \Bbb Q$.
By Banach fixed point theorem, there exists unique fixed point $a$ of $f$. 
Let $g(x) = x - f(x)$. Note that $a$ is a fixed point of $f$ if and only if $g(a) = 0$, and that $f(\Bbb Q)\subseteq \Bbb Q$ if and only if $g(\Bbb Q)\subseteq \Bbb Q$. Since you are looking for $f$ such that $f(x)\neq x,\ \forall x\in\Bbb Q$, the unique fixed point of $f$ must be irrational, i.e., $g$ has unique irrational zero.
We can now rewrite $(1)$ as follows:
\begin{align}
|f(x)-f(y)|\leq \frac 12|x-y| &\iff |g(x)-g(y)-(x-y)|\leq\frac 12|x-y|\\
&\iff \left|\frac{g(x)-g(y)}{x-y}-1\right|\leq\frac 12\\
&\iff \frac 12\leq \frac{g(x)-g(y)}{x-y}\leq \frac 32\tag{2}
\end{align}
Relation $(2)$ can be used to derive simple sufficient condition on $g$ to get $f$ with desired properties: 

Let $g\colon\Bbb R\to\Bbb R$ be differentiable such that $\frac 12\leq
 g'(x)\leq \frac 32, \forall x\in\Bbb R,$ and let $g$ have irrational
   zero. Then $f(x)=x-g(x)$ (restricted on $\Bbb Q$) satisfies condition
   $(1)$ and has no rational fixed point. If $g(\Bbb Q)\subseteq\Bbb Q$,
   then $f(\Bbb Q)\subseteq\Bbb Q$ as well.

Proof. From the above discussion, all that remains to be proven is that $\frac 12\leq g'(x)\leq \frac 32$ implies $(2)$. Assume that there exist some $x,y\in\Bbb R$ such that $$\frac{g(x)-g(y)}{x-y}\not\in\left[\frac 12,\frac 32\right].$$ By the mean value theorem, there exists some $c\in\Bbb R$ such that $g'(c)\not\in \left[\frac 12,\frac 32\right]$. Contradiction. $\tag*{$\square$}$
Using this, you should be able to come up with as many examples as you'd like, but let me just discuss the one Ian suggested in the comments. It comes from Newton's method  for numerical approximation of root of $x^2-2$. So, we could let $g(x) = \frac{x^2-2}{2x}$, but there are some troubles with it. First of all, it is not defined at $0$. Secondly, $g'(x) = \frac 12 + \frac 1{x^2}$ and $g'(x)\in\left[ \frac 12,\frac 32\right]$ if and only if $x^2\geq 1$. This leads to considering changing definition of $g$ on $(-\infty,1)$. We want to do it in simplest way possible so that $g$ remains differntiable and $g'(x)\in\left[ \frac 12,\frac 32\right]$. Since $g(1) = -\frac 12$ and $g'(1)=\frac 32$, we consider function $x\mapsto\frac 32 x - 2$, i.e. we can define
$$g(x) =\left\{
\begin{array}{c l}
\frac{x^2-2}{2x}, &x\geq 1\\
\frac 32 x-2, &x<1
\end{array}\right.$$ which is differentiable and satisfies $g'(x)\in\left[ \frac 12,\frac 32\right]$ on whole $\Bbb R$. This gives your $f$.
