Defining an Embedding in the Product Topology $f: X \to X \times Y$ and $f(x) = x \times y_0$ for $y_0 \in Y$ Munkres defines a topological embedding as follows.

Now suppose that $f:X\to Y$ is an injective continuous map,where $X$ and $Y$ are topological spaces. Let $Z$ be the image set $f(X)$, considered as a subspace of $Y$; then the function $f':X\to Z$ obtained by restricting the range of $f$ is bijective. If $f'$
happens to be a homeomorphism of $X$ with $Z$, we say that the map $f:X\to Y $ is a topological embedding, or simply an embedding, of $X$ in $Y$.

I met some confusion once, I began studying the problem

Given $y_0 \in Y$ prove that $f: X \to X \times Y$ defined by $f(x) = x \times y_0$ is a toplogical embedding.

I am attempting to prove that $f$ restricted to $Z=\{x \times y_0 : x \in X\}$ is a homeomorphism to prove that the entire mapping is an embedding. I have already shown that $f$ is bijective by verifying that it is both injective and surjective. Furthermore, I proved that $f$ is continuous by showing that if $V\subset Y$ is an open set, then $f^{-1}(V) \subset X$ is also an open set.
My confusion lies in demonstrating that $f^{-1}$ restricted to $Z$ is also a continuous map. For this end, I am trying to prove that if $U\subset X$ is an open set, then $f(U)\subset Y$ is also an open set. My confusion now centers around the phrasing of Munkres question. I have recalled the definition o the product Topology as being generated by the basis $\mathcal{B}_{X \times Y}$ defined by
$$\mathcal{B}_{X \times Y} = \{U \times V: U \in \mathcal{T}_X, V \in \mathcal{T}_Y\}$$
where $\mathcal{T}_X$ and $\mathcal{T}_Y$ are the respective topologies on $X$ and $Y$. Then, a set is open in the product topology $\mathcal{T}_{X\times Y}$ if and only if it may be expressed as a union of sets in $\mathcal{B}_{X\times Y}$. Singleton sets (and finite sets) are closed in Hausdorff spaces for example, so how may I be certain that $\{y_0\}$ and $U \times \{y_0\}$ for $U \subset X$ be open in the product topology. Here lies my confusion about the definition of a topological embedding. When Munkres says to consider $Z$ as a subspace of $Y$ would I be allowed to neglect $\{y_0\}$? To prove that $f$ is continuous, I simply examine two cases depending on whether or not $\{y_0\}$ was open. If it was not open, then $f$ would be continuous in the vacuous sense, whereas if $\{y_0\}$ was open I would proceed in the canonical way.
Problem: May someone please elaborate on Munkres definition of an embedding and the particular considerations necessary for applying it to a product topology.
Note: My question is not a duplicate of this as I am more concerned with the concepts behind general definitions than their application to this problem.
 A: Munkres' definition seems to be quite standard. The idea of an embedding between $X$ and $Y$ is just a homeomorphism between $X$ and a subspace of $Y$. Usually the expression "there is a copy of $X$ in $Y$" is used. It's cool because this allows to treat $X$ as a subspace of $Y$.
Let's apply it to your particular problem of (finite) product spaces. What you want to show is that $X$ is embedded in $X\times Y$, in particular, that $x\mapsto (x, y_0)$ is an embedding. First, some intuition: Chances are that you have already used this notion when you have said (at some point) that $\mathbb{R}\subseteq \mathbb{C}$. Remember that $\mathbb{C}=\mathbb{R}^2$, the only difference being it's algebraic structure (in this case that doesn't apply, we are doing general topology ignoring algebra). So $\mathbb{R}\subseteq \mathbb{C}$ is equivalent to saying $\mathbb{R}\subseteq \mathbb{R}^2$. In this case, what you really mean is $\mathbb{R}\times\{y_0\}\subseteq \mathbb{R}^2$ and that $\mathbb{R}$ is equivalent in some sense to $\mathbb{R}\times \{y_0\}$ (in this case the similarity is the homeomorphism, the question at hand). In summery: one could say that the real line is imbedded in the plane.
Let's formalize this:  Let $\prod_\alpha Y_\alpha$ be an arbitrary product, and $y_0=({y_{0}}_{\alpha})$ a given point. For each index $\beta$, the set
$$S(y_0;\beta)=Y_\beta\times \prod\{{y_0}_\alpha\mid \alpha\neq \beta\}\subseteq \prod_\alpha Y_\alpha$$
is called the slice in $\prod_\alpha Y_\alpha$ through $y_0$ parallel to $Y_\beta$. In the previous example, $S((y_1,y_2);1)=\mathbb{R}\times \{y_2\}$ and $S((y_1,y_2);2)=\{y_1\}\times\mathbb{R}$. In your particular problem, $S((x_0,y_0);1)=X\times \{y_0\}=\{x\times y_0\mid x\in X\}$ (where $x_0$ is any $x_0\in X$, it doesn't matter).
It's a well known property that

Theorem. The map $s_\beta\colon Y_\beta\to S(y_0;\beta)\subseteq \prod_\alpha Y_\alpha$ given by
$$y_\beta\mapsto y_\beta\times \prod\{{y_0}_\alpha\mid \alpha\neq \beta\}$$
is a homeomorphism of $Y_\beta$ with the subspace $S(y_0;\beta)$. That is, $s_\beta$ is an embedding of $Y_\beta$ into $\prod_\alpha Y_\alpha$.

That's the general theorem in regards of embeddings of a space of the product into the product space. The proof can be found, among other places, in Dugundji's Topology Chap. IV theorem 3.2.
For the "particular considerations you may want to have when doing embeddings into product spaces", are the same you'd have when showing homeomorphisms of product spaces, it's not particularly special. I suggest reading the proof of the theorem mentioned above, it's a good example you can guide yourself with.
I won't go any further since you already have the solution to your problem in the link you mention in your question and in Dugundji's.
