3 homeomorphisms between spaces (2 with jungle metric) I've been studying for my final exam from topology and I found such an exercise.

Let $X=([0,1]\times\{0\})\cup \bigcup_{n=1}^{\infty}(\{\frac{1}{n}\}\times[0,\frac{1}{n}])$
Let $Y=([0,1]\times\{0\})\cup \bigcup_{n=1}^{\infty}(\{\frac{1}{n}\}\times[0,1])$

*

*Are $(X,\rho_e)$ and $(Y,\rho_e)$ homeomorphic?

*Are $(X,\rho_r)$ and $(X,\rho_e)$ homeomorphic? ($\rho_r$ is the jungle river metric)

*Are $(Y,\rho_r)$ and $(Y,\rho_e)$ homeomorphic?


What I think:
If we draw $X$ we get a horizontal line $[0,1]$ with vertical lines which are smaller if we approach $(0,0)$.
If we draw $Y$ we get a similar picture but the vertical lines are of the same height.
I think that  $(X,\rho_e)$ and $(Y,\rho_e)$ are homeomorphic. The horizontal line will be the same horizontal line and we can transform the vertical lines by scaling.
I know it is not a very formal proof, but that is my intuition. I don't know what to do with the other homeomorhpisms. I tried to show that $X$ is not compact or connected in the jungle metric, but I got no results. (I think that $X$ and $Y$ are arc-conected in the euclidean metric so they should be also connected).
What do you think? Could you help me with this problem?
Edit:
Jungle river metric, where $x=(x_1,x_2)$, $y=(y_1,y_2)$:
$\rho_r(x,y)=\rho_e(x,y)$ if $x_1=y_1$
$\rho_r(x,y)=|x_2|+|x_1-y_1|+|y_2|$ if $x_1\neq y_1$
 A: $X$ and $Y$ are not homeomorphic in the euclidean metric: $X$ is compact, while $Y$ is not. You can see that $Y$ is not compact because it is not closed in $\mathbb {R}^2:$ the sequence $(1/n,1)$ is a sequence in $Y$ that converges to $(0,1),$ which is not in $Y.$ Now $Y'=Y\cup (\{0\}\times [0,1])$ is compact; it equals $\{0,1,1/2, \dots\}\times [0,1],$ the cross product of two compact sets. Since $X = Y' \cap \{(x,y): y\le x\},$ the intersection of a compact set with a closed set, $X$ is compact.
A: HINT: For the first question, show that $X$ is a closed, bounded subset of the plane in the Euclidean metric, while $Y$ is not. What topological property does this give $X$ but not $Y$?
For the other two, show that at every point $x$ of $X$ or $Y$ that is not $\langle 0,0\rangle$ there is an $\epsilon_x>0$ such that $B_{\rho_r}(x,s)=B_{\rho_e}(x,s)$ whenever $s<\epsilon_x$. Then show that $\langle 0,0\rangle$ has the same open nbhds in $\langle X,\rho_r\rangle$ and $\langle X,\rho_e\rangle$, and the same open nbhds in 
$\langle Y,\rho_r\rangle$ and $\langle Y,\rho_e\rangle$.
