Let $X$ be the set of all finite sequences $\langle x_0,\ldots,x_n\rangle$ such that $n\in\Bbb N$, and $x_k\in\Bbb R^+$ for $k=0,\ldots,n$. For $r>0$ and $x=\langle x_0,\ldots,x_n\rangle\in X$ let $B(x,r)$ be the set of points $y=\langle y_0,\ldots,y_m\rangle\in X$ such that
- $m\ge n$;
- $y_k=x_k$ for $k<n$;
- $|y_n-x_n|<r$; and
- $|y_k|<r$ for $n<k\le m$.
Then $\{B(x,r):x\in X\text{ and }0<r\in\Bbb R\}$ is a base for a connected topology $\tau$ on $X$. In fact, for each $\langle x_0,\ldots,x_n\rangle\in X$ the subspace
$$\{\langle x_0,\ldots,x_{n-1},x\rangle\in X:x\in\Bbb R^+\}$$
is homeomorphic to $\Bbb R^+$ with the usual topology, so $X$ is even path connected.
Now let $x=\langle x_0,\ldots,x_n\rangle\in X$, and let $Y=X\setminus\{x\}$. Then $Y$ has the following three components:
$C_0(x)=\{\langle y_0,\ldots,y_m\rangle\in X:m\ge n\text{ and }y_n>x_n\text{ and }y_k=x_k\text{ for }k=0,\ldots,n-1\}$
$C_1(x)=\{\langle y_0,\ldots,y_m\rangle\in X:m>n\text{ and }y_k=x_k\text{ for }k=0,\ldots,n\}$
$C_2(x)=Y\setminus\big(C_0(x)\cup C_1(x)\big)$
The intuitive idea is straightforward and is essentially the same as that of Kaj Hansen’s answer. We start with $\Bbb R^+$; points on it correspond to the sequences in $X$ of length $1$. To each point $x\in\Bbb R^+$ we attach another copy of $[0,\to)=\{0\}\cup\Bbb R^+$ by identifying the $0$ of $[0,\to)$ with $x$; the sequence $\langle x,y\rangle\in X$ then corresponds to the point $y$ on the copy of $[0,\to)$ attached to $x$. The topology is that induced by the so-called jungle river metric on $\Bbb R^+\times\Bbb R^+$.
Then we do it again: to each point $\langle x,y\rangle$ we attach a copy of $[0,\to)$ by identifying $\langle x,y\rangle$ with the $0$ of that copy of $[0,\to)$, and we extend the topology in an analogous fashion. We keep going to get sequences of arbitrary finite length. The space $X$ is the direct limit of the finite stages of this construction.