$(A^ctimes B)cup(Atimes B^c)cup(A^c times B^c)=(A^ctimes Y)cup(Xtimes B^c)$

Already in the first step you’ve written something that suggests a misconception on your part. It’s true that $(Atimes B^c)cup(A^ctimes B^c)=(Acup A^c)times(B^ccup B^c)$, but only because $(Atimes B^c)cup(A^ctimes B^c)=(Acup A^c)times B^c$; it is not in general true that

$$(Wtimes X)cup(Ytimes Z)=(Wcup Y)times(Xcup Z);.tag{1}$$

For instance, if $W=X={0}$ and $Y=Z={1}$, then

$$begin{align*} (Wtimes X)cup(Ytimes Z)&=({0}times{0})cup({1}times{1})\ &={langle 0,0rangle}cup{langle 1,1rangle}\ &={langle 0,0rangle,langle 1,1rangle};, end{align*}$$

but

$$begin{align*} (Wcup Y)times(Xcup Z)&=({0}cup{1})times({0}cup{1})\ &={0,1}times{0,1}\ &={langle 0,0rangle,langle 0,1rangle,langle 1,0rangle,langle 1,1rangle};. end{align*}$$

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 $$(Xtimes Z)cup(Ytimes Z)=(Xcup Y)times Z;;tag{2}$$ you should prove this, if you’ve not done so already. Thus, a good first couple of steps would be

$$begin{align*} (A^ctimes B)cup(Atimes B^c)cup(A^ctimes B^c)&=(A^ctimes B)cupbig((Acup A^c)times B^cbig)\ &=(A^ctimes B)cup(Xtimes B^c);. end{align*}$$

Clearly $(A^ctimes B)cup(Xtimes B^c)subseteq(A^ctimes Y)cup(Xtimes B^c)$, so you could finish the proof by showing that $(A^ctimes Y)cup(Xtimes B^c)subseteq(A^ctimes B)cup(Xtimes B^c)$.

There are several ways to do this; for instance, you could use ‘element-chasing’, assuming that $langle x,yranglein(A^ctimes Y)cup(Xtimes B^c)$ and showing that $langle x,yranglein(A^ctimes B)cup(Xtimes B^c)$. In keeping with the more algebraic style that you were using, however, you could notice that

$$A^ctimes Y=A^ctimes(Bcup B^c)=(A^ctimes B)cup(A^ctimes B^c)$$

by $(2)$, so that

$$begin{align*} (A^ctimes Y)cup(Xtimes B^c)&=(A^ctimes B)cup(A^ctimes B^c)cup(Xtimes B^c)\ &=(A^ctimes B)cup(Xtimes B^c);, end{align*}$$

since $A^ctimes B^csubseteq Xtimes B^c$.

And as you can see, we’ve not just shown that $$(A^ctimes Y)cup(Xtimes B^c)subseteq(A^ctimes B)cup(Xtimes B^c);:$$ we’ve shown that $$(A^ctimes Y)cup(Xtimes B^c)=(A^ctimes B)cup(Xtimes B^c);.$$

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:

$$begin{align*} (A^ctimes B)cup(Atimes B^c)cup(A^ctimes B^c)&=(A^ctimes B)cupbig((Acup A^c)times B^cbig)\ &=(A^ctimes B)cup(Xtimes B^c)\ &=(A^ctimes B)cupbig((A^ctimes B^c)cup(Xtimes B^c)big)\ &=big((A^ctimes B)cup(A^ctimes B^c)big)cup(Xtimes B^c)\ &=(A^ctimes Y)cup(Xtimes B^c);. end{align*}$$