Proof by Contradiction relating to rational and irrational numbers

It's correct. A bit too lengthy, though. From $$ 2x-\frac{m}{n}=\frac{p}{q} $$ you can derive $$ 2x=\frac{m}{n}+\frac{p}{q}=\frac{mq+np}{nq} $$ so $$ x=\frac{mq+np}{2nq} $$ would be rational, because you assumed $m,n,p,q\in\mathbb{Z}$ with $n\ne0$ and $q\ne0$, so $mq+np\in\mathbb{Z}$ and $2nq\ne0$.

Since $x+y$ is rational, if $x - y$ is rational, then so are $(x + y) - (x - y)$ and $(x + y) + (x - y)$

From your "$2x -\frac{m}{n} = \frac{p}{q}$, you should realize you have "twice an irrational is the sum of two rationals": "$2x = \frac{p}{q} + \frac{m}{n}$. The sum of two rationals is rational, twice an irrational is irrational, and you have your contradiction at your third line.