I would specify that $p,q,m,n\in\mathbb{Z},q\neq0,n\neq0$, in your first line of the proof.
Other than that, the rest looks good.
" $$ a + br =...= (pn + qmr) / qn $$
...
Since r is irrational, we know that both the numerator and the denominator cannot be rational numbers.
"
I think your conclusion is illogical / not deductive.
Your assumption was:
"Assume that if a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is rational."
The numerator, $pn + qmr, $ is in the form $c+dr$ where $c$ and $d$ are rational and $r$ is irrational. How did you deduce that the numerator, $pn + qmr, $ is irrational, without assuming the thing you're trying to prove?
In fact, your assumption implies that the numerator $is$ rational.
You were on the right track in your proof until this part:
$a + br = p/q + (m/n)r/1$
I think you got "caught up in the maths" and forgot about the logical reasoning of the proof.
I would slightly modify the first line of your proof, which I assume you intended to be a proof by contradiction:
Proof: Assume that if a and b are rational numbers, b ≠ 0, and r is an irrational number, $and \ that \ $ a + br is rational. (Now your goal is to prove that some sort of contradiction will arise.)
By the definition of rational, we can substitute a and b with fractions where p, q, m, n are particular but arbitrary integers. (this bit is fine)
Your assumption assumed $a+br \ $ is rational, so now you should write:
$a + br = p/q$
and take it from there. Remember your goal now is to get a contradiction based on the fact that r is irrational and the rest of the numbers are rational.
Best Answer
The proof is correct. Note that the proof uses only the property that the rationals are closed under multiplication $\,\color{#c00}{\rm(CM)}\,$ so it will work more generally for any such subset $\,R\,$ of "numbers", i.e. $$\smash[t]{\, a\in R,\ ab \not\in R\,\Rightarrow\, b\not\in R,\ \ {\rm else} \ \ a,b\in R\,\stackrel{\rm\color{#c00}{(CM)}}\Rightarrow\, ab\in R,\ \ \ {\rm contra\ hypothesis.}}$$
Hence $\,a\,$ integral , $\,ab\,$ nonintegral $\,\Rightarrow\, b\,$ nonintegral
and $\,a\,$ algebraic , $\,ab\,$ transcendental $\,\Rightarrow\, b\,$ transcendental
and $\,a\,$ constant, $\,ab\,$ nonconstant $\,\Rightarrow\, b\,$ nonconstant, $ $ for functions, polynomials, etc.
The statement is just a contrapositive form of the fact that $\,R\,$ is closed under multiplication, i.e. writing out the logical equivalance explicitly, using $\ x\Rightarrow y\, \equiv\, \lnot x\ \ {\rm or}\ \ y,\,$ we have
$$\begin{eqnarray} && a\in R\ {\rm and}\ ab\not\in R\,\Rightarrow\ b\not\in R \\ \equiv\, && a\not\in R\ \ \,or\ \ ab\in R\ \ {\rm or}\ \ b\not\in R \\ \equiv\, && a\not\in R\ \ \,or\ \ b\not\in R\ \ \,{\rm or}\ \ ab\in R \\ \equiv\, && a\in R\ {\rm and}\ b\in R\,\Rightarrow\ ab\in R \\ \end{eqnarray}$$