Is the factor of a topological product closed in the product topology? I could not find an answer to the question of whether a factor of a topological product is closed in the product topology itself, so I wrote my own proof. My questions are under the proof itself.
Let $X$ and $Y$ be two topological Hausdorff spaces.
Let $X\times Y$ be their product equipped with the product topology.
To show:
Let $y\in Y$ arbitrary, then $X\times \left\lbrace y\right\rbrace$ is a closed subset in $X\times Y$ and $X\times \left\lbrace y\right\rbrace$ is homeomorphic to $X$.
Proof:
Let $A\subseteq X$ and $B\subseteq Y$ be closed, such that:
$A = A^\prime\backslash X$ and $B = B^\prime\backslash Y$, where $A^\prime$ and $B^\prime$ are open.
Then we find with elementary set operations as can be found e.g. here https://en.wikipedia.org/wiki/List_of_set_identities_and_relations that:
$\begin{align}
A \times B &= \left(X\backslash A^\prime\right) \times \left(Y\backslash B^\prime\right)\\
           &= \left( X \times Y\right) \backslash \left(\left(A^\prime \times Y \right) \cup \left(X \times B^\prime\right)\right).
\end{align}$
Thereby we see that $A \times B$ is the complement of an open set in $X \times Y$ and thus closed in $X \times Y$.
Applying this to $X\times \left\lbrace y\right\rbrace$ yields:
$\begin{align}
X\times \left\lbrace y\right\rbrace &= \left(X \backslash \emptyset\right) \times \left\lbrace y\right\rbrace\\
&= \left(X \backslash \emptyset\right) \times \left(Y\backslash\left(Y\backslash\left\lbrace y\right\rbrace\right)\right)\\
&=\left( X \times Y \right) \backslash \left[\left(\emptyset \times Y \right)\cup \left(X \times \left(Y \backslash \left\lbrace y\right\rbrace\right)\right)\right]\\
&=\left( X \times Y \right) \backslash \left[\emptyset \cup \left(X \times \left(Y \backslash \left\lbrace y\right\rbrace\right)\right)\right]\\
&=\left( X \times Y \right) \backslash \left[ X \times \left(Y \backslash \left\lbrace y\right\rbrace\right)\right].
\end{align}$
Thus, the complement of $X\times \left\lbrace y\right\rbrace$ in $X \times Y$ is $X \times \left(Y \backslash \left\lbrace y\right\rbrace\right)$.
$X \times \left(Y \backslash \left\lbrace y\right\rbrace\right)$ is open in $X \times Y$ because $X$ is open in $X$ and $Y \backslash \left\lbrace y\right\rbrace$ is open in $Y$ since $Y$ is a Hausdorff space and thus a $T_1$ space in which every singleton set is closed. 
Thus, we have shown that $X\times \left\lbrace y\right\rbrace$ is closed in $X \times Y$ equipped with the product topology proving the first statement.
Now for the second statement:
$X\times \left\lbrace y\right\rbrace$ is open in the subspace topology because if $W \subseteq Y$ is an open neighbourhood of $y$ (which always exists), then $X\times \left\lbrace y\right\rbrace = \left( X\times \left\lbrace y\right\rbrace\right) \cap \left( X \times W\right)$ and $X \times W$ is open in $X \times Y$.
Therefore, we immediately see that the open sets in $X\times \left\lbrace y\right\rbrace$ are all of the form $U\times \left\lbrace y\right\rbrace$, where $U \subseteq X$ open.
This allows us to construct a homeomorphism $\widetilde{\pi}: X\times \left\lbrace y\right\rbrace \to X$ which is simply the projection on the first component.
An inverse is given by $\widetilde{\pi}^{-1}: X\to X\times \left\lbrace y\right\rbrace$ which functions in the obvious way by simply "multiplying" an element in $X$ with $\left\lbrace y\right\rbrace$.
One also easily sees that $\widetilde{\pi}$ is continuous because if $U\subseteq X$ is open then so is its preimage under $\widetilde{\pi}$ which is $U\times \left\lbrace y\right\rbrace$ in $X\times \left\lbrace y\right\rbrace$.
Thus $\widetilde{\pi}$ is a homeomorphism and we find that $X \cong X\times \left\lbrace y\right\rbrace$. This proves our claim.
q.e.d.
Since I am pretty new to topology I have a few questions concerning my proof especially its second part:
1.) Is it correct? If not, where did I go wrong and how?
2.) Do $X$ and $Y$ need to be Hausdorff or is it sufficient to require they be $T_1$-spaces?
Thanks in advance and cheers!
Edit: Fixed a typo.
 A: If $Y$ is a $T_1$ space, and $y\in Y$, then $\{y\}$ is a closed set, and therefore $Y\setminus\{y\}$ is an open set. So, $X\times(Y\setminus\{y\})$ is open, and therefore its complement is closed. But $\bigl(X\times(Y\setminus\{y\})\bigr)^\complement=X\times\{y\}$. I think that this is simpler than your proof.
And I don't know the meaning of “$X\times\{y\}$  is open in the subspace topology”. It is open as a subset of which space? Not in $X\times Y$, I hope, since that's not true in general.
But, yes, it looks as if your proof of the fact that $X$ and $X\times\{y\}$ are homeomorphic is correct.
A: Q1:
Your first proof is correct, but you argue a bit redundantly: You show that $A \times B$ is always closed if $A,B$ are closed. Thus there is no need to write it down once more for $X×\{y\}$ provided $\{y \}$ is closed (as it is in $T_1$-spaces).
In your second proof it is unclear what you mean by "$X×\{y\}$ is open in the subspace topology." It is in general not an open subspace of $X \times Y$, and if you want to say is is open in the subspace $X×\{y\}$, then it is trivial.
But whatever it means, you do need it in the rest of your proof which is correct.
Q2:
You only need to know that $\{y\}$ is closed in $Y$ to show that $X×\{y\}$ is closed in $X \times Y$ (no conditions on $X$). Thus $Y$ being $T_1$ is enough. And for the homeomorphism part you do not need any condition at all. Look at your proof!
