The theorem tells us that $\Bbb R^n$ is homogeneous: for any points $a,b\in\Bbb R^n$ there is a surjective homeomorphism $h_{a,b}:\Bbb R^n\to\Bbb R^n$ carrying $a$ to $b$, i.e., such that $h_{a,b}(a)=b$. In general $h_{a,b}\ne h_{c,d}$ if $\langle a,b\rangle\ne\langle c,d\rangle$.
Intuitively speaking, a homogeneous space is one in which all points look the same: there is no topological property that distinguishes one point from another. The circle $S^1$ is another homogeneous space: for any $a,b\in S^1$ we can take $h_{a,b}$ to be any map that rotates the circle about its centre through an angle that takes $a$ to $b$.
A less obvious example of a homogeneous space is the middle-thirds Cantor set $C$. Fix $a,b\in C$; $a$ and $b$ have ternary expansions containing no $1$s, so we can write
$a=\sum_{n\ge 1}\frac{a_n}{3^n}\quad\text{ and }\quad b=\sum_{n\ge 1}\frac{b_n}{3^n}\;,$
where $a_n,b_n\in\{0,2\}$ for $n\in\Bbb Z^+$. Define $h_{a,b}:C\to C$ as follows. If $c=\sum_{n\ge 1}\frac{c_n}{3^n}$ with $c_n\in\{0,2\}$ for $n\in\Bbb Z^+$, let
$\widehat{c_n}=\begin{cases}c_n,&\text{if }a_n=b_n\\2-c_n,&\text{if }a_n\ne b_n\;,\end{cases}$
and let $h_{a,b}(c)=\sum_{n\ge 1}\frac{\widehat{c_n}}{3^n}\;.$
Clearly $h_{a,b}(a)=b$, and it’s a nice little exercise to verify that $h_{a,b}$ is a surjective homeomorphism. (You might find this a little surprising, since at first glance it might appear that the endpoints of the deleted middle thirds are somehow different from the other points of $C$.)
On the other hand, the closed unit interval $[0,1]$ is not homogeneous: there is no surjective homeomorphism $h:[0,1]\to[0,1]$ taking either of the endpoints to a point in $(0,1)$. This is because the endpoints are distinguished from the other points by the following topological property: if $x\in(0,1)$, then $[0,1]\setminus\{x\}$ is not connected, but if $x\in\{0,1\}$, then $[0,1]\setminus\{x\}$ is connected. (The points of $(0,1)$ are said to be cut points of $[0,1]$.) As another exercise you might try to show that the property of being a cut point is preserved by homeomorphisms.
Even more extreme types of inhomogeneity are possible. Let $X_0$ be a closed line segment in $\Bbb R^2$ of length $1$, and let $x_0$ be its midpoint. Form $X_1$ by attaching a segment of length $1/2$ to $x_0$ that meets $X_0$ only at $x_0$. Let $x_1,x_2$, and $x_3$ be the midpoints of the ‘arms’ of $X_1$. Form $X_2$ by attaching two segments to $x_1$, three to $x_2$, and four to $x_3$ in such a way that the new ‘arms’ have length $\le1/4$, are pairwise disjoint, and do not intersect $X_1$ except at their points of attachment. Continue in this fashion: given $X_n$ with last point of attachment $x_m$ and $k$ ‘arms’, let $x_{m+1},\dots,x_{m+k}$ be the midpoints of those arms, and for $i=1,\dots,k$ attach $m+i+1$ segments of length $\le 2^{-k}$ to $x_{m+i}$ in such a way that the new ‘arms’ are pairwise disjoint and meet $X_n$ only at their respective points of attachment. The resulting set is $X_{n+1}$.
Now let $X=\bigcup_{n\in\Bbb N}X_n$, and let $D=\{x_n:n\in\Bbb N\}$. $X$ inherits its topology and metric from $\Bbb R^2$, and it’s not hard to check that $D$ is a dense subset of $X$. For each $n\in\Bbb N$, $X\setminus\{x_n\}$ has $n+2$ components. The number of connected components of $X\setminus\{x\}$ is a topological property of $x$: it’s preserved by homeomorphisms. Thus, no homeomorphism $h:X\to X$ can take $x_m$ to $x_n$ if $m\ne n$. Moreover, if $x\in X\setminus D$, then $X\setminus\{x\}$ has two components, so no homeomorphism $h:X\to X$ can take any point of $D$ to a point of $X\setminus D$. Thus, if $h:X\to X$ is a homeomorphism, $h(x_n)=x_n$ for every $n\in\Bbb N$. Now use the fact that $h$ is continuous and $D$ is dense in $X$ to show that $h(x)=x$ for every $x\in X$. In other words, the identity map is the only homeomorphism of $X$ to itself: no two distinct points of $X$ ‘look alike’ in a topological sense.