A particular proof of the existence of infinitely many right-inverses Let $R$ be an infinite ring and $a$ be an element with two different right-inverses. Then $a$ has infinitely many right-inverses.
Here and here we have proofs which I'm able to follow. My initial idea, though, was to take the distinct right-inverses $r_1$ and $r_2$ of $a$, and parametrize the line segment by $\gamma:R\to R$ given by $\gamma(x)=r_1x+r_2 (1-x)$. Then $a\gamma(x)=1$ for all $x\in R$. But I'm having trouble checking that the image of $\gamma$ must be infinite. So:

Questions: Must the image of $\gamma$ really be infinite? Is this idea fixable?

 A: Here is a counterexample.  Let $k$ be a finite field, let $V=k^{\oplus\mathbb{N}}$, and let $W$ be a nonzero finite-dimensional vector space over $k$.  Let $R$ be the ring of endomorphisms $f:V\oplus W\to V\oplus W$ such that $f(V)\subseteq V$.  Let $a\in R$ be the endomorphism which is the left shift operator on $V$ and the identity on $W$.  Let $b\in R$ be the endomorphism which is the right shift operator on $V$ and the identity on $W$.  Let $d\in R$ be a nonzero endomorphism which is $0$ on $V$ and maps all of $W$ to the kernel of $a$ (i.e., the subspace of $V$ supported on the first coordinate).
Now observe that $ab=1$ and $ad=0$, so $r_2=b$ and $r_1=b+d$ are two different right inverses to $a$.  For any $x\in R$, we then have $\gamma(x)=b+dx$, so the question is whether $dx$ takes infinitely many values.  But any element of $R$ of the form $dx$ must annihilate $V$ and have image contained in the kernel of $a$, since $x$ maps $V$ to itself and $d$ annihilates $V$ and has image contained in the kernel of $a$.  Since $W$ and the kernel of $a$ are both finite, there are only finitely many such maps.  So $dx$ takes only finitely many different values, and so does $\gamma(x)$.

To relate your approach to the standard proof, note that if you fix one right inverse $r_2=b$, then you can get another right inverse $r_1=b+d$ by adding any $d\in R$ such that $ad=0$.  One particular choice of such a $d$ is $d=1-ba$.  The standard proof then shows that the elements $da^n$ are distinct for all $n\in\mathbb{N}$, so that $\gamma(x)=b+dx$ takes infinitely many different values.  This only works, though, because of the special value of $d$ chosen (in particular, the argument uses the fact that $1-d\in Ra$).  As the example above shows, if you pick an arbitrary nonzero $d$ such that $ad=0$, there may be only finitely many different values of $dx$.
