Is the $Cl(A \times B)=Cl(A) \times Cl(B)$. Here is my the part of my proof that I have questions about namely; the first containment.
$
\subset)
$
$$
(x,y)\in Cl(A \times B)
$$
Therefore every closed set in the product topology looks like
$$
(x,y) \in \bigcap_\alpha( U_\alpha \times V_\alpha)|\text{$U_\alpha$  closed in $A$ and $V_\alpha$ closed in B }
$$
So now is the part I am unsure about is the rest
$$
x\in\bigcap_\alpha U_\alpha \quad \quad y\in \bigcap_\alpha V_\alpha
$$
Thus
$$
x\in Cl(A) \quad y\in Cl(B)
$$
Therefore
$$
(x,y)\in Cl(A)\times Cl(B).
$$
 A: I assume that $A,B$ are subsets of spaces $X,Y$? But your proof is not correct, since it does not use the (correct) definition of the closure.
Here is a quick proof using the characterization of $cl(A)$ as the set of points of $X$  whose open neighborhoods intersect $A$.
Hence, $cl(A) \times cl(B)$ is the set of points $(x,y) \in X \times Y$ whose open neighborhoods of the form $U \times V$, where $U$ is an open neighborhood of $x$ and $V$ is an open neighborhood of $y$, intersect $A \times B$. But since every open neighborhood of $(x,y)$ is a union of such $U \times V$, the claim follows.
A: Here's how you might show the equality using limits in case the space is metric (or, more generally, first-countable).

Fact 1: $x \in \mathrm{Cl}(X)$ iff there is a sequence $x_1, x_2, \dots \in X$, such that $x_n \rightarrow x$.
Fact 2: $(x_n, y_n) \rightarrow (x,y)$ iff $x_n\rightarrow x$ and $y_n\rightarrow y$.

Let $(a,b) \in \mathrm{Cl}(A\times B)$. Then there is a sequence $(a_1,b_1), (a_2, b_2), \dots \in A\times B$ such that $(a_n,b_n) \rightarrow (a,b)$. Then $a_1, a_2, \dots \in A$ and $a_n\rightarrow a$, and therefore $a \in \mathrm{Cl}(A)$. Likewise, $b \in \mathrm{Cl}(B)$. So $(a,b) \in \mathrm{Cl}(A)\times\mathrm{Cl}(B)$.
Conversely, let $(a,b) \in \mathrm{Cl}(A)\times\mathrm{Cl}(B)$. Since $a \in \mathrm{Cl}(A)$, there is a sequence $a_1, a_2, \dots \in A$, such that $a_n \rightarrow a$. Likewise, there's a sequence $b_1, b_2, \dots \in B$, such that $b_n \rightarrow b$. Then $(a_1, b_1), (a_2, b_2), \dots \in A\times B$ and $(a_n, b_n) \rightarrow (a,b)$. So $(a,b) \in \mathrm{Cl}(A\times B)$.

Here's an analogous version for a general topological space using nets.

Fact 1: $x \in \mathrm{Cl}(X)$ iff there is a net $(x_d)_{d\in D}$ in $X$, such that $\lim_{d \in D} x_d \rightarrow x$.
Fact 2: Let $(x_d)_{d \in D}$ be a net in $X$, and let $(y_e)_{e \in E}$ be a net in $Y$, such that $\lim_{d \in D} x_d = x \in X$ and $\lim_{e \in E} y_e = y \in Y$. Then $D\times E$ is a directed set with $(d,e)\leq(d',e') \iff (d\leq d'\ \wedge\ e\leq e')$, and $\lim_{(d,e) \in D\times E} (x_d, y_e)  \rightarrow (x,y)$.
Fact 3: Let $\big((x_d, y_d)\big)_{d \in D}$ be a net in $X\times Y$ such that $\lim_{d \in D} (x_d, y_d) = (x,y) \in X\times Y$. Then $\lim_{d\in D}x_d = x$ and $\lim_{d \in D}y_d = y$.

Let $(a,b) \in \mathrm{Cl}(A\times B)$. Then there is a net $\big((a_d,b_d)\big)_{d \in D}$ in $A\times B$ such that $\lim_{d\in D} (a_d,b_d) \rightarrow (a,b)$. Then $\lim_{d \in D} a_d = a$, and therefore $a \in \mathrm{Cl}(A)$. Likewise, $b \in \mathrm{Cl}(B)$. So $(a,b) \in \mathrm{Cl}(A)\times\mathrm{Cl}(B)$.
Conversely, let $(a,b) \in \mathrm{Cl}(A)\times\mathrm{Cl}(B)$. Since $a \in \mathrm{Cl}(A)$, there is a net $(a_d)_{d\in D}$ in $A$, such that $\lim_{d \in D}a_d \rightarrow a$. Likewise, there's a net $(b_e)_{e \in E}$ in $B$, such that $\lim_{e\in E}b_e \rightarrow b$. Then $\lim_{(d,e) \in D\times E}(a_d, b_e) \rightarrow (a,b)$. So $(a,b) \in \mathrm{Cl}(A\times B)$.
