Consider a random walk in which each step is either $\binom{-1}{-1}$ or $\binom{1}{-1}$.
Define "Furthest-at-Finish path" (abbreviated as FF) as a path in which the maximum horizontal distance from the origin is reached only at the end.
So for example,
The path $\binom{1}{-1}, \binom{-1}{-1}, \binom{1}{-1}, \binom{1}{-1}$ is an FF because the maximum horizontal distance from the origin, $2$, is reached only at the end.
The path $\binom{1}{-1}, \binom{1}{-1}, \binom{-1}{-1}$ is not an FF because the maximum horizontal distance from the origin, $2$, is reached before the end.
Prove that the number of FF with $2k$ steps is $\binom{2k}{k}$.
I have a proof, shown below. I am looking for a simpler, more intuitive proof.
There does exist a bijective proof, using something called the Catalan factorization of a random walk. This proof is given in Random walks and Catalan factorizations, by Ömer Eğecioğlu and Alastair King. The authors show how to give a bijection from $FF$ paths, to the set of paths from $A$ to $A'$ of length $2k$, where $A'$ is the point $2k$ units south of $A$. Obviously, the number of paths $A\to A'$ is $\binom{2k}k$. However, their bijection is not something I would call intuitive.
To equate your setup to the one used in the paper, you need to rotate your picture $90^\circ$ clockwise. That is, instead of counting walks from $A$ to $C$ whose steps are $\binom{-1}{-1}$ and $\binom{-1}1$, you consider walks from $C$ to $A$ when the picture is rotated $90^\circ$ anti-clockwise, using steps $\binom{1}{1}$ and $\binom{1}{-1}$. Now, the condition $FF$ in the original setup translates to saying that the walk never dips below the horizontal line passing through $C$ (which is the green line in your picture, rotated $90^\circ$). In the language of the paper, these are nonnegative walks of length $2k$. By proposition $2$ in that paper, they are in bijection with balanced walks of length $2k$, which are obviously enumerated by $\binom{2k}k$.