To close this question, let me present a more explicit answer than the one given in the comments.

First, let $w:[0,+\infty)\to [0,+\infty)$ be constructed as follows. On $[1,2]$ define $w$ to be the tent map, with $w(1)=w(2)=0$, and $w(3/2)=4$. Then copy-paste scaled versions of this tent infinitely in both sides so that the graph consists of a sequence of congruent triangles.

It is proven in Silverman - On Maps with Dense Orbits and the Definition of Chaos, p 360, that this map has a property that for every open non-empty $U\subset [0,+\infty)$ there is

$k\in\mathbb{Z}$ and $m\in\mathbb{N}$ such that $[0,2^{k+2r}]\subset w^{2r}(U)$, for $r>m$.

Now consider $v:[0,+\infty)^n\to [0,+\infty)^n$ defined by $v(x_1,...,x_n)=(w(x_1),...,w(x_n))$. Let $U\subset [0,+\infty)^n$ be open and non-empty. There are $U_i\subset [0,+\infty)$, such that $U_1\times...\times U_n\subset U$. From the property of $w$ there are $k_1,...,k_n\in\mathbb{Z}$ and $m\in\mathbb{N}$ such that $[0,2^{k_1+2r}]\times...\times[0,2^{k_m+2r}] \subset v^{2r}(U_1\times...\times U_n)\subset v^{2r}(U)$, for $r>m$.

Now observe that $[0,+\infty)^n$ is homeomorphic to $W=[0,+\infty)\times(-\infty,+\infty) ^{n-1}$ with the boundary corresponding to the boundary. Below I will view $v$ as a map on $W$. Then the set $\{(x_1,...,x_n), x_1=0\}$ is invariant with respect to $v$, and also if $U\subset W$ is open and nonempty we have $\bigcup v^{r}(U)= W$.

Let $h:\mathbb{R}^n\to \mathbb{R}^n$ be the reflection $f(x_1,...,x_n)=(-x_1,x_2,...,x_n)$. Define $f:\mathbb{R}^n\to \mathbb{R}^n$ by

$$f\left(x_1,...,x_n\right)=\left\{\begin{array}{ll} h(v(x_1,...,x_n)) & x_1\ge 0 \\ v(h(x_1,...,x_n)) & x_1\le 0
\end{array}\right.$$

Note that when $x_1=0$, $h$ does nothing, but also since $v$ maps boundary into the boundary, the first coordinate of $v(x_1,...,x_n)$ is $0$, from where the map is well-defined, and therefore continuous. Observe that $f^2|_{W}=v^2$, and so if $U\subset \mathbb{R}^n$ is open and nonempty one can show that $\bigcup f^{r}(U)= \mathbb{R}^n$, which implies TT.