Does every sphere defined by jungle/paris metric is a retract of $\mathbb{R}^2$ with jungle/paris topology? In this case, firstly I'm being told that in ($\mathbb{R}^n$, $d_e$), where $d_e$ is the euclidean metric, every closed sphere defined by this metric $D^n(x_0, x)=\{x\in \mathbb{R}^n:\ d(x_0, r) \le r\}$, is a retract of $\mathbb R^n$.
We define retraction as a $f: ((X, T_x) \to (A, T_A))$, where A $\subset$ X and $T_A$ is a topology of $A \cap U$, $U \in T_x$ such that $\forall a\in A \ \ f(a) = a$, and then $A$ is a retract of $X$.
The questions is if every sphere with jungle/paris metric is a retract of ($\mathbb{R}^2$, $T_j$) or ($\mathbb{R}^2$, $T_p$), where $T_j$, $T_p$ are respectively a jungle/paris topology (i.e. $\mathbb{R}^2$ with jungle/paris metric).
We define jungle metric as $d_r(A, B)=\begin{cases} d_e(A, B),  \ \ \  \ \ \mbox{when A, B are on the same orthogonal to X axis line;} \\
d_e(A , A_1) + d_e(A_1, B_2) + d_e(B_2, B), \ \ \ \  \mbox{in every other case.} \end{cases}$, 
where $A_1$, $B_2$ are projections of $A$ and $B$ respectively on $X$ axis and $d_e$ is euclidean metric,
and paris metric as a:


*

*Euclidean distance when the two points belong to a line through the origin (0,0)

*otherwise the sum of the distances of the two points from the origin. 


here is how the spheres with a real-valued r and some $x_0$ look like:

So I've found two functions that possibly can be treated like retractions on these spaces, but I really think I'm missing something and there is an obvious counterargument to this statement (like if there existed such a set in one or maybe both of these topologies that doesn't satisfy the rule).
What I have is:
$1)$ retraction is a surjection and a quotient map $r: ((X, T_x) \to (A, T_A))$ and therefore images of closed sets are closed in $(A, T_A)$ and naturally images of open sets are open in $(A, T_A))$, and also inverse images of closed sets are closed in $(X, T_x)$ and inverse images of open sets are open in $(X, T_x)$.
$2)$ two functions: 
$f: (\mathbb{R}^2 \to D^2(x_0, r)) = 
\begin{cases} 
x,  \ \ \ \ \ \mbox{when $d_p(x_0 , x)\le r$;} \\
(0,0)  \ \ \ \ \ \mbox{when $d_p(x_0, x) > r$.} \end{cases}$
$f: (\mathbb{R}^2 \to D^2(x_0, r)) = 
\begin{cases} 
x,  \ \ \ \ \ \mbox {when $d_j(x_0 , x)\le r$;} \\
x_0,  \   \mbox{when $d_j(x_0, x) > r$.} \end{cases}$
where $d_j$ is a jungle metric, $d_p$ a paris one, and $D^2(x_0, r)$ a sphere defined by equivalent metric with the beginning in $x_0$ and a radius r.
3) Both ($\mathbb{R}^2$, $T_j$) and ($\mathbb{R}^2$, $T_p$) are Hausdorff because every metric space is Hausdorff and thus if there exists a retract of any of these spaces, it must be a close set of $\mathbb{R}^2$. 
Could anyone help with a proper counterargument?
 A: Note that everywhere that you’ve written $D^n(x_0,x)$ or $D^2(x_0,x)$ you should have $D^n(x_0,r)$ or $D^2(x_0,r)$, respectively. Your function for the Paris metric doesn’t necessarily map $\Bbb R^2$ to $D^2(x_0,r)$, because the origin need not be in $D^2(x_0,r)$. Your $f$ for the jungle metric is not continuous: the inverse image of a small enough open nbhd of a point on the boundary of $D^2(x_0,r)$ is not open.
For the Paris metric it’s helpful to use polar coordinates. Each $x\in\Bbb R^2$ except the origin can be represented in polar coordinates as $\langle r,\theta\rangle$ for a unique $r>0$ and $\theta\in[0,2\pi)$. For $\theta\in[0,2\pi)$ let $L_\theta=\{\langle r,\theta\rangle:r\ge 0\}$. If $x_0=\langle r_0,\theta_0\rangle$ is a point other than the origin, and $\epsilon\le r_0$, then $D^2(x_0,\epsilon)$ is just the closed line segment from $\langle r_0-\epsilon,\theta_0\rangle$ to $\langle r_0+\epsilon,\theta_0\rangle$ on $L_{\theta_0}$. Show that the map $$\Bbb R^2\to L_{\theta_0}:\langle r,\theta\rangle\mapsto\langle r,\theta_0\rangle$$ is a retraction; then find a retraction of $L_{\theta_0}$ onto $$D^2(x_0,\epsilon)=\{\langle r,\theta_0\rangle\in L_{\theta_0}:r_0-\epsilon\le r\le r_0+\epsilon\}\;.$$ (HINT: Find a retraction of the closed half-line $[0,\to)$ onto the closed interval $[1,2]$.)
If $x_0$ is the origin, $D^2(x_0,\epsilon)$ is the same set of points in the Paris metric as it is in the Euclidean metric, and collapsing everything radially to its boundary still works.
The jungle metric also has two different kinds of basic open sets. If $x_0=\langle a,b\rangle$ with $b\ne 0$, and $\epsilon\le|b|$, then $D^2(x_0,\epsilon)$ is a closed segment on the line $x=a$: $$D^2(x_0,\epsilon)=\{a\}\times[b-\epsilon,b+\epsilon]\;.$$
In this case a two-step retraction again works well: first show that the map
$$\Bbb R^2\to\{a\}\times\Bbb R:\langle x,y\rangle\mapsto\langle a,y\rangle$$
is a retraction, then retract $\{a\}\times\Bbb R$ onto $D^2(x_0,\epsilon)$.
If $b=0$, on the other hand, $D^2(x_0,\epsilon)$ is a square diamond, as shown in your diagram. In this case try projecting vertically to the boundary of $D^2(x_0,\epsilon)$ for $a-\epsilon\le x\le a+\epsilon$ and sending $\langle x,y\rangle$ to $\langle a+\epsilon,0\rangle$ for $x\ge a+\epsilon$ and to $\langle a-\epsilon,0\rangle$ for $x<a-\epsilon$.
In both metrics there are disks of mixed type: in the Paris metric when $x_0$ is not the origin and $\epsilon>r_0$, and in the jungle metric when $b\ne 0$ and $\epsilon>|b|$. Once you sort out the simple disks, these shouldn’t pose too much more trouble.
