Why $(X,d)$ is a complete $\mathbb{R}$-tree? 
Definition. An $\mathbb{R}$-tree is a metric space $(X,d)$ such that
  
  
*
  
*there is a unique geodesic segment (denoted $[x,y]$) joining each pair of points $x,y \in X$;
  
*if $[x,y] \cap [y,z] = \{y\}$, then $[x,y] \cup [y,z] = [x,z]$.
  
  
  
Example. Consider the following equivalence relation $\mathcal{R}$ on the set $\mathbb{N} \times [0,1]$:
  $$(n,0) \mathcal{R} (m,0)\text{ and }(n,t)\mathcal{R}(n,t)\quad(n,m\in\mathbb{N}, t\in[0,1]).$$ Now, define the following metric on $X := ( \mathbb{N} \times[0,1] ) / \mathcal{R}$:
  $$
d((n,t),(n,s)) = |t-s|\text{ and }d((n,t),(m,s))=t+s,\text{ when }n \neq m.
$$
  It is easy to see that $(X,d)$ is a complete $\mathbb{R}$-tree and satisfies the ($\overline{Q_4}$) condition.

Why $(X,d)$ is a complete $\mathbb{R}$-tree?
 A: The space under consideration is the (metric) hedgehog space of spininess $\aleph_0$. It is more easily visualised as countably many copies of the unit interval $[0,1]$ with their origins identified. For each $n \in \mathbb{N}$ we'll call the subset $\{n\} \times (0,1]$ the $n$th spine of $X$, and denote the identified point by $\mathbf{0}$.
For the most part, this question is just filling in details.
First we handle the geodesics.  I'll describe them, but will omit the verifications. Suppose that $\mathbf{x}, \mathbf{y}$ are distinct points of $X$.


*

*If $\mathbf{x}$ and $\mathbf{y}$ both belong to the $n$th spine of $x$ (so that $\mathbf{x} = \langle n,x \rangle$ and $\mathbf{y} = \langle n,y \rangle$ for some $x,y \in (0,1]$, then the geodesic from $\mathbf{x}$ to $\mathbf{y}$, $[\mathbf{x},\mathbf{y}] = \alpha : [0,1] \to X$, is defined by $$\alpha (t) = \langle n , (t-1)x + ty \rangle.$$

*If $\mathbf{x}$ and $\mathbf{y}$ belong to different spines of $x$ (say that $\mathbf{x} = \langle n,x \rangle$ and $\mathbf{y} = \langle m,y \rangle$ for some $x,y \in (0,1]$ and distinct $n,m \in \mathbb{N}$), then the geodesic from $\mathbf{x}$ to $\mathbf{y}$, $[\mathbf{x},\mathbf{y}] = \alpha : [0,1] \to X$, is defined by $$\alpha (t) = \begin{cases}
\langle n , x-(x+y)t \rangle, &\text{if }0 \leq t < \frac{x}{x+y} \\
\mathbf{0}, &\text{if }t = \frac{x}{x+y} \\
\langle m , (x+y)t-x \rangle, &\text{if }\frac{x}{x+y} < t \leq 1.
\end{cases}$$

*If $\mathbf{x} = \mathbf{0}$ and $\mathbf{y}$ belongs to the $n$th spine of $X$ (so that $\mathbf{y} = \langle n,y \rangle$ for some $y \in (0,1]$), then the geodesic from $\mathbf{x}$ to $\mathbf{y}$, $[\mathbf{x},\mathbf{y}] = \alpha : [0,1] \to X$, is defined by $$\alpha (t) = 
\begin{cases}
\mathbf{0}, &\text{if }t = 0\\
\langle n, yt \rangle, &\text{if }0 < t \leq 1.
\end{cases}$$

*If $\mathbf{x}$ belongs to the $n$th spine (so that $\mathbf{x} = \langle n,x \rangle$ for some $x \in (0,1]$) and $\mathbf{y} = \mathbf{0}$, then the geodesic from $\mathbf{x}$ to $\mathbf{y}$, $[\mathbf{x},\mathbf{y}] = \alpha : [0,1] \to X$, is defined by $$\alpha (t) = 
\begin{cases}
\langle n,(1-t)y \rangle, &\text{if }0 \leq t < 1 \\
\mathbf{0}, &\text{if }t = 1.
\end{cases}$$
From here it is relatively easy to determine what the images of these geodesics are as subsets of $X$, and then to show that they satisfy the required property.
For example, suppose $\mathbf{x} = \langle n,x \rangle$, $\mathbf{y} = \langle n,y \rangle$, $\mathbf{z} = \langle m,z \rangle$ where $m \neq n$, then the images of the geodesics $[\mathbf{x},\mathbf{y}]$, $[\mathbf{y},\mathbf{z}]$, respectively, are 
$$
\{ n \} \times [ \min\{x,y\} , \max\{x,y\} ];
\quad
( \{ n \} \times (0,y] ) \cup \{\mathbf{0}\} \cup ( \{m\} \times (0,z] ).
$$
Also, the image of the geodesic $[\mathbf{x},\mathbf{z}]$ is
$$
( \{ n \} \times (0,x] ) \cup \{\mathbf{0}\} \cup ( \{m\} \times (0,z] ).
$$
If $[\mathbf{x},\mathbf{y}] \cap [\mathbf{y},\mathbf{z}] = \{ \mathbf{y} \}$ then it must be that $y \leq x$ (why?), and therefore $[\mathbf{x},\mathbf{z}] = [\mathbf{x},\mathbf{y}] \cup [\mathbf{y},\mathbf{z}]$.
