I have some question that I get for homework.
Let $U$ and $W$ subspace of Euclidean space $E$. Then if $U$ is orthogonal on $W$, then $U^{\bot}$ is orthogonal on $W^{\bot}$.
For second question I know that if $U$ is orthogonal on $W$ than for every $x\in U$ and for every $y\in W$ $(x|y)=0$, but I do not know is this true, but I think like this, that because $U$ is orthogonal on $W$ than $dim U \geq dim W$(I never read this so I am not so sure is this true) but we have that $dim U+dim U^{\bot}=dimE$ and $dimW+dimW^{\bot}=dimE$ if use both than $dimU^{\bot}\le dimW^{\bot}$ so $W^{\bot}$ is orthogonal on $U^{\bot}$, so it is not true,but I do not sure.
What you are trying to prove is wrong. $U \bot W$ doesn't necessarily implies that $U^\bot \bot W^\bot$.
You can find an exemple of a situation where it doesn't stand in the comments, in my discution with mfl.
But just to keep a more general idea, we'll keep a generic Euclidian space $E$.
Let's just assume that $DimE \ge 2$, you can check the cases $DimE = 0 $ and $ DimE=1$, independantly if you need them.
Consider U={$0_E$}, where $0_E$ is the null vector of $E$.
We have $DimU=0$ and therefore, $DimU^\bot=DimE-0=DimE$.
Since $U^\bot \subset E$ and $DimU^\bot=DimE$, we have $U^\bot=E$
Let's consider $W$ a non-trivial subspace of $E$ ($DimE \gt DimW \ge 1$).
By definition of the scalar product, $\forall x \in W, (0_E,x) = 0$
So $U \bot W$
Now $DimW^\bot=DimE-DimW\ge1$.
So $\exists x_{w^\bot} \in W^\bot, x_{w^\bot} \ne 0_E$
Since $W^\bot \subset E$, $x_{w^\bot}\in E = U^\bot$
So, let's put $x=y=x_{w^\bot}$:
$$\exists x \in W^\bot, y \in \ U^\bot, (x|y)= (x_{w^\bot}|x_{w^\bot})=||x_{w^\bot}||² > 0 $$