Elementary Proof Relating to Pythagorean Triples

I think your proof seems to work fine. However, maybe it’s worth mentioning when you state that ‘an integer multiple of an irrational number remains irrational’, you are excluding the case where that integer is zero. Obvious, but worth mentioning for rigour. Perhaps state at the beginning that $a,b,c$ are positive integers.

The only other thing I think you could make clearer is how you know that $c>a$ and $c>b$. The triangle inequality actually says that $a + b \geq c$ (or $a+b>c$ if you exclude degenerate triangles), so perhaps you should give a few more steps to show how you arrive at your conclusion.

I would say that you're proof is fine!

There's only a claim that I would change.

$c>a$ and $c>b\:$ by the triangle inequality.

However, the triangle inequality simply states that the inequality $$a+b>c$$ is true for all its permutations...

I would write instead

$$c=\sqrt{a^2+b^2}>\sqrt{a^2+0}=a\iff c>a$$

---

You may also want to be more rigorous and prove the claim

Since an integer [nonzero-]multiple of an irrational number remains irrational...

You can, for instance, use contradiction again.

Let $x\in\Bbb N, y\in\Bbb R\setminus \Bbb Q$ and suppose that for some $z\in\Bbb Q$ such that $z=\frac pq$ (recall the definition of rational) $$x\cdot y=z\iff y=\frac{p}{qx}$$ which is a contradiction, since irrational numbers cannot be expressed as a fraction between to integers...