GCD proof - by contradiction?

Let $c$ be any common divisor of $a+b$ and $a-b$. Then $c$ divides $(a+b)+(a-b)$ and $(a+b)-(a-b)$. So $c$ divides $2a$ and $2b$.

Note that there are integers $x$ and $y$ such that $ax+by=1$ and therefore $(2a)x+(2b)y=2$. It follows that $c$ divides $2$. Thus no number greater than $2$ can divide both $a+b$ and $a-b$.

For completeness, you should show that each of the possibilities $\gcd(a+b,a-b)=1$ and $\gcd(a+b,a-b)=2$ can happen.

Hint $\,\ {\rm mod}\ a\!-\!b\!:\,\ a\equiv b\,\Rightarrow\, \color{#0a0}{a\!+\!b\equiv 2b}\,$ so, by the Euclidean Algorithm $\rm\color{#c00}{(EA)}$

$\qquad (a\!-\!b,\,\color{#0a0}{a\!+\!b})\overset{\rm\color{#c00}{EA}} = (a\!-\!b,\,\color{#0a0}{2b}) = (a\!-\!b,\,2),\ $ by $\ (a\!-\!b,b)\overset{\rm\color{#c00}{EA}}= (a,b)=1,\,$ and Euclid's Lemma.

Here $\rm\color{#c00}{EA}$ means $\ (m,n)\, =\, (m,n')\ $ if $\ n'\!\equiv n\pmod m\,\ $ [= descent step of Euclidean Algorithm].