$(A^c\times B)\cup(A\times B^c)\cup(A^c \times B^c)=(A^c\times Y)\cup(X\times B^c)$ Let $(X,T)$ and $(Y,T')$ be two topological spaces and $A\subset X ,B\subset Y$.Show that $$(A^c\times B)\cup(A\times B^c)\cup(A^c \times B^c)=(A^c\times Y)\cup(X\times B^c)$$
solution $$1...........(A^c\times B)\cup(A\times B^c)\cup(A^c \times B^c)=(A^c\times B)\cup\Biggl((A\cup A^c)\times(B^c\cup B^c)\Biggl)$$
$$2........=(A^c\times B)\cup(X\times (B^c\cup B^c))$$
$$3.......=(A^c\cup X \times B\cup (B^c\cup B^c))$$
$$4......=(A^c\cup X \times (B\cup B^c)\cup B^c)$$
$$5......=(A^c\cup X \times Y\cup B^c)$$
$$6....=(A^c\times Y)\cup(X\times B^c)$$
Is this proof correct?

Already in the first step you’ve written something that suggests a misconception on your part. It’s true that $(A\times B^c)\cup(A^c\times B^c)=(A\cup A^c)\times(B^c\cup B^c)$, but only because $(A\times B^c)\cup(A^c\times B^c)=(A\cup A^c)\times B^c$; it is not in general true that
$$(W\times X)\cup(Y\times Z)=(W\cup Y)\times(X\cup Z)\;.\tag{1}$$
For instance, if $W=X=\{0\}$ and $Y=Z=\{1\}$, then
$$\begin{align*}
(W\times X)\cup(Y\times Z)&=(\{0\}\times\{0\})\cup(\{1\}\times\{1\})\\
&=\{\langle 0,0\rangle\}\cup\{\langle 1,1\rangle\}\\
&=\{\langle 0,0\rangle,\langle 1,1\rangle\}\;,
\end{align*}$$
but
$$\begin{align*}
(W\cup Y)\times(X\cup Z)&=(\{0\}\cup\{1\})\times(\{0\}\cup\{1\})\\
&=\{0,1\}\times\{0,1\}\\
&=\{\langle 0,0\rangle,\langle 0,1\rangle,\langle 1,0\rangle,\langle 1,1\rangle\}\;.
\end{align*}$$
The remaining steps are ambiguous, since you didn’t use enough parentheses, but it appears that you really did think that $(1)$ is true and used it at your step $3$. There it definitely fails.
What is true is that $$(X\times Z)\cup(Y\times Z)=(X\cup Y)\times Z\;;\tag{2}$$ you should prove this, if you’ve not done so already. Thus, a good first couple of steps would be
$$\begin{align*}
(A^c\times B)\cup(A\times B^c)\cup(A^c\times B^c)&=(A^c\times B)\cup\big((A\cup A^c)\times B^c\big)\\
&=(A^c\times B)\cup(X\times B^c)\;.
\end{align*}$$
Clearly $(A^c\times B)\cup(X\times B^c)\subseteq(A^c\times Y)\cup(X\times B^c)$, so you could finish the proof by showing that $(A^c\times Y)\cup(X\times B^c)\subseteq(A^c\times B)\cup(X\times B^c)$.
There are several ways to do this; for instance, you could use ‘element-chasing’, assuming that $\langle x,y\rangle\in(A^c\times Y)\cup(X\times B^c)$ and showing that $\langle x,y\rangle\in(A^c\times B)\cup(X\times B^c)$. In keeping with the more algebraic style that you were using, however, you could notice that
$$A^c\times Y=A^c\times(B\cup B^c)=(A^c\times B)\cup(A^c\times B^c)$$
by $(2)$, so that
$$\begin{align*}
(A^c\times Y)\cup(X\times B^c)&=(A^c\times B)\cup(A^c\times B^c)\cup(X\times B^c)\\
&=(A^c\times B)\cup(X\times B^c)\;,
\end{align*}$$
since $A^c\times B^c\subseteq X\times B^c$.
And as you can see, we’ve not just shown that $$(A^c\times Y)\cup(X\times B^c)\subseteq(A^c\times B)\cup(X\times B^c)\;:$$ we’ve shown that $$(A^c\times Y)\cup(X\times B^c)=(A^c\times B)\cup(X\times B^c)\;.$$
If you wanted to, you could reverse the order of this last calculation and append it to the first part to get a direct proof of equality:
$$\begin{align*}
(A^c\times B)\cup(A\times B^c)\cup(A^c\times B^c)&=(A^c\times B)\cup\big((A\cup A^c)\times B^c\big)\\
&=(A^c\times B)\cup(X\times B^c)\\
&=(A^c\times B)\cup\big((A^c\times B^c)\cup(X\times B^c)\big)\\
&=\big((A^c\times B)\cup(A^c\times B^c)\big)\cup(X\times B^c)\\
&=(A^c\times Y)\cup(X\times B^c)\;.
\end{align*}$$
