Is the closed strip homeomorphic to the closed half-plane? Let
$$
X = \{(x,y) \in \mathbb{R}^2: 0 \le x \le 1\}
\quad 
Y = \{(x,y) \in \mathbb{R}^2: 0 \le x\}.
$$
Are $X$ and $Y$ homeomorphic?
I first thought no because their boundaries
are not homeomorphic however the argument is flawed
because the theorem I want to use is:
If $X \cong Y$ then for $A \subseteq X$
$$
\partial_X (A) \cong \partial_Y(f(A)).
$$
So applying this theorem would yield
$$
\partial_X(X) \cong
\partial_Y(Y)
\implies \emptyset \cong \emptyset
$$
which is not helpful. Any hints?
 A: To proof it without using any algebraic topology, the same as here works:
Call a space $X$ compactly connected, if
for each compact subset $A$ of $X$ there is compact subset $B$ of $X$, such that $A \subseteq B$ and $X \setminus B$ is connected.

$Y = [0,\infty) \times \mathbb{R}$ is compactly connected, since each compact subset is contained in a compact rectangle, such that one of its edges lies on the $y$-axis. The remainder consists of three (infinite, non-closed) "rectangles", where one intersects the other two. Hence it is connected.

$X = [0,1] \times \mathbb{R}$ is not compactly connected:
Let $A := [0,1] \times \{0\}$. Then for any compact $B$ with $A \subset B$, $X \setminus B$ is not connected.
A: Since
$$
X = [0,1] \times \mathbb{R} \cong [0,1] \times (0,1)
\quad 
Y = [0,\infty) \times \mathbb{R} \cong [0,1) \times
(0,1)
$$
it suffices to show that the latter are not
homeomorphic and this has been asked in
Homeomorphisms in $\mathbb{R}^2$.
I believe the accepted answer is incorrect because of the
same reason I mentioned, taking boundaries in $\mathbb{R}^2$
rather than on $X$ and on $Y$, however the comment to
that answer appears to be correct (defining the "boundary" as the set of points so that $X \setminus \{p\}$ is simply connected).
