Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$
But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq \frac{1}{df}$. So $\frac{1}{bf} + \frac{1}{df} \leq \frac{1}{bd}$. Multiply both sides by $bdf$ and we get that $b+d\leq f$.
First of all, the proof is correct and I congratulate you on the excellent effort. I will only offer a few small comments on the writing.
It's not clear until all the way down at (5) that you intend to do a proof by contradiction, and even then you never make it explicit. It's generally polite to state at the very beginning of a proof if you plan to make use of contradiction, contrapositive, or induction.
Tiny detail, maybe even a typo: $n$ and $m$ are integers, not necessarily naturals, so the statement at the end of (2) needs to reflect that. But for integers, also $x>-1$ implies $x\geq 0$, so it's not a big deal.
You didn't really need to make $n$ and $m$ positive since the only place you use positivity is at the very, very end you need $m>0$ to derive the contradiction. You don't even use it when you multiply by $d$ since that relied on the expressions being positive and not the individual numbers themselves. This is the only place I can imagine really see simplifying the proof.
As it stands, it would make the reader more comfortable if you named the positive versions of $(n,m)$ as $(N,M)$ or $(n',m')$ or something.
Finally, as you hint at, you don't need to consider the positive-positive case. But perhaps you should be more explicit why this is, earlier in the proof.
Best Answer
You do it the same way we do with numbers: find a common denominator and then just add the numerators.
For example, if you needed to add $$\frac{1}{2}+\frac{1}{3}$$ you need to find a "common denominator" (a number which is a multiple of both $2$ and $3$), convert the two fractions to the common denominator, and then add them. The simplest common denominator is just the product of the two denominators, we one way to do this is: $$\begin{align*} \frac{1}{2}+\frac{1}{3} &= \frac{1}{2}\cdot\frac{3}{3} + \frac{1}{3}\cdot\frac{2}{2}\\ &= \frac{3}{6} + \frac{2}{6}\\ &= \frac{3+2}{6}\\ &= \frac{5}{6}.\end{align*}$$
So do the same thing here: $$\begin{align*} \frac{4x+4h}{x+h+1} - \frac{4x}{x+1} &= \frac{4x+4h}{x+h+1}\cdot\frac{x+1}{x+1} + \frac{4x}{x+1}\cdot\frac{x+h+1}{x+h+1} \\ &= \frac{(4x+4h)(x+1)}{(x+h+1)(x+1)} - \frac{4x(x+h+1)}{(x+h+1)(x+1)}.\end{align*}$$ Now you can just add them, since they have the same denominator, by keeping the denominator and adding numerators. Do the algebra in the numerator, and simplify as needed.
Note: If you are more comfortable with the algebra, there are things you can do to simplify your life; when adding fractions, sometimes the product of the denominators is not the best common denominator: if you can spot a simpler one, use it. For instance, if we had $\frac{1}{6}+\frac{1}{10}$, then we could use $30$ as the common denominator, instead of $6\times 10 = 60$, which leads to smaller numbers, easier arithmetic; likewise, if we had $$\frac{1}{x^2-1} + \frac{2}{x^2+2x+1}$$ then you could use $(x^2-1)(x^2+2x+1)$ as a common denominator, but if you could spot that $x^2-1 = (x+1)(x-1)$ and $x^2+2x+1 = (x+1)^2$, then you might realize that you can use $(x+1)^2(x-1)$ instead, which is smaller (degree 3 instead of degree 4).
In your situation, on possible simplification is to note that both fractions are multiples of $4$. So you can factor out $4$ and deal with simpler fractions if you want. But if you are unsure about these kinds of simplifying steps, then don't do them. They are not strictly necessary, though they are useful when dealing with more complicated expressions.