$mI have to demonstrate
$$m<n \land n<r\Rightarrow  m<r$$
where $m,n,r \in \mathbb Z$ using order definition of an integer.
Attempt:
$$\\m<n\hspace{.3cm}if \hspace{.3cm}m \le n \hspace{.3cm}and \hspace{.3cm}m\ne0\rightarrow n=m+p^+
\\n<r \hspace{.3cm}if \hspace{.3cm}n \le r\hspace{.3cm}and \hspace{.3cm}n\ne0\rightarrow r=n+q^+
\\r=m+p^++q^+
\\r=m+(p^++q)^+$$ 
I don't know how to continue.
 A: As the first answer says, you are almost there already.  All that is required is to make explicit the key step that $\;p^+ + q^+\;$ is again a positive integer.
Just to give you another viewpoint, here is a slightly different proof in a slightly different style.  As in the question, all variables range over the integers.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$From the solution attempt in your question, I'm guessing that your "order definition of an integer" is as follows (in slightly different notation):
$$
\tag 0 m < n \;\equiv\; \langle \exists p : p > 0 : n = m + p \rangle
$$
So that is what I will be using here.

First we notice that $\ref 0$ is equivalent to the simpler
$$
\tag 1 m < n \;\equiv\; n - m > 0
$$ as we see by the following simple calculation:
$$\calc
\langle \exists p : p > 0 : n = m + p \rangle
\calcop\equiv{arithmetic}
\langle \exists p : p > 0 : p = n - m \rangle
\calcop\equiv{logic: one-point rule}
n - m > 0
\endcalc$$
Now, let's start on the most complex side of our demonstrandum, and work towards the other side:
$$\calc
m < n \;\land\; n < r
\calcop\equiv{definition $\ref 1$, twice}
n - m > 0 \;\land\; r - n > 0
\calcop{\tag{*} \then}{arithmetic: the sum of two positive number is positive}
(n - m) + (r - n) > 0
\calcop\equiv{arithmetic: simplify}
r - m > 0
\calcop\equiv{definition $\ref 1$}
m < r
\endcalc$$
This completes the proof.

This proof makes explicit the key step $\ref *$ of this proof: positive integers add up to a positive integer.
A: You've done all the work, just notice the last step. 
If $n > m$ then there exists $p^+ \in \mathbb Z$ such that $n = m + p^+$, there also exists $q^+$ such that $r = n + q^+$ thus 
$$r = m + \underbrace{(p^+ + q^+)}_{=s^+\in \mathbb Z} = m + \color{#f05}{s^+} \implies r > m$$
