Non nesting action on Real Trees Background
Let $G$ be a group acting by homeomorphisms on an $\mathbf{R}$-tree $T$. The element $ g \in G $ is elliptic if the fixed point set $\operatorname{Fix}(g)$ is non-empty. The group action is non-nesting if no $ g \in G $ maps an arc properly into itself.
Claim
Let $ g \in G $ be elliptic. If $ x \not \in \operatorname{Fix}(g)$, then $\operatorname{Fix}(g) \cap  [x, gx]$ consists of precisely one point.
Available Proof
Given $x \not \in \operatorname{Fix}(g)$, choose $ x_0 \in \operatorname{Fix}(g)$, and defne $u$ by $ [x_0, x] \cap \operatorname{Fix}(g) = [x_0, u]$. Non-nesting implies $\operatorname{Fix}(g) \cap [x, gx] = \{ u \} $.
My doubt
Why does Non-nesting imply that $Fix g  \cap [x, gx] = \{ u \} $ works?

First, the property of being at most a singleton is true without assumption on $g$:

Let $T$ be an $\mathbf{R}$-tree. Then for an arbitrary self-homeomorphism $g$ of $T$ and $x\in T$, the intersection $\mathrm{Fix}(g)\cap [x,gx]$ is at most 1 point.

Indeed, if $y,z$ are fixed points in $[x,gx]$, say with $y\in [x,z]$, then $y=gy\in [gx,gz]=[z,gx]$, which forces $y=z$.
Next:

If $g$ is elliptic and both $g$ and $g^{-1}$ are non-nesting then this is indeed a singleton.

Let $x_0$ be a fixed point. Let $v$ be the median point of the triple $\{x_0,x,gx\}$. Since both $g$ and $g^{-1}$ are non-nesting, none of $[x_0,v]$ and $[x_0,gv]$ is properly contained in the other. Hence, $gv=v$ and $v\in [x,gx]$. 
