$X \times \{y\}$ is homeomorphic to $X$ Can I get a proof verification? The only thing I am unsure about is the proof of continuity of $f,f^{-1}$. I know the proof of bijectiveness is trivial.
Prove:$X \times \{y\}$ is homeomorphic to $X$.
Attempt: Take the fact that we can prove continuity by showing that the inverse image of basis elements are open. Define the map $f:X \times \{y\} \rightarrow X$ by $f(x,y)=x$. Then $f$ is one to one since $f(x,y)=f(w,y)\implies x=w$ and so $(x,y)=(w,y)$. $f$ is surjective, since for any $x \in X, f(x,y)=x$. So $f$ is bijective. Let $U$ be an open set in $X$. Then $f^{-1}(U)=U \times \{y\}=(X \times \{y\}) \cap (U \times Y)$, which is open in the subspace topology, so $f$ is continuous.Let $W$ be any open set in $X \times \{y\}$. Then $W=(U \times V) \cap (X \times \{y\})$ where $U \times V$ is open in the product topology. Then $f(W)=U$ is open in $X$ and so $f^{-1}$ is continuous.

Most of it is fine, but the proof that $f$ is an open set needs a bit more work. The problem is that it isn’t immediately obvious that an open set $W$ in $X\times\{y\}$ is of the form $(U\times V)\cap(X\times\{y\}$ for some open $U\subseteq X$ and $V\subseteq Y$: all you really know is that $W=U\cap(X\times\{y\})$ for some open subset $U$ of $X\times Y$, and that $U$ need not be ‘rectangular’.
Here’s one way to get around the problem. Let $U$ be such a set. For each $x\in f[W]$ there are open $G_x\subseteq X$ and $V_x\subseteq y$ such that
$$\langle x,y\rangle\subseteq G_x\times V_x\subseteq U\,.$$
Let $G=\bigcup_{x\in f[W]}G_x$; $G$ is open in $X$. Let $z\in G$. Then $z\in G_x$ for some $x\in f[W]$, so $\langle z,y\rangle\in G_x\times V_x\subseteq U$. Clearly $\langle z,y\rangle\in X\times\{y\}$, so $\langle z,y\rangle\in W$. I’ll leave it to you to verify that if $z\in X\setminus G$, then $\langle z,y\rangle\notin W$ and conclude that $f[W]=G$ and hence that $f$ is open.
