$W^{\bot}$ is a subspace of $U^{\bot}$? 
Let $U$ and $W$ be subspaces of an inner product space $V$. If $U$ is a subspace
  of $W$, then $W^{\bot}$  is a subspace of $U^{\bot}$?.

I don't find the above statement intuitively obvious. Could someone provide a proof?
 A: If you're orthogonal to everything in a set, then you're also orthogonal to everything in every subset of that set.
Put another way: Elements of $W^\perp$ have to be orthogonal to more vectors than elements of $U^\perp$; they have to be orthogonal to the vectors in $U$ and the vectors in $W\setminus U$ (if any).  Therefore there are less of them than if we took the vectors that only have to satisfy the property of being orthogonal to vectors in $U$.  (More restrictions$\implies$ Fewer vectors satisfying the restrictions.)  
A: It should be intuitive, already at the level of logic:
To be in $W^\perp$, you have to satisfy a certain condition $P(w)$ (namely: 'be orthogonal to $w$') for each and every element $w\in W$. 
So given a subset $U\subseteq W$, to be in $U^\perp$ means you have to satisfy $P(w)$ merely for all $u\in U$. 
Thus you have to satisfy less properties to be in $U^\perp$, thus it is easier to be in $U^\perp$, thus $U^\perp$ is larger: $W^\perp\subseteq U^\perp$.
It should also be intuitive geometrically: consider $\mathbb{R}^3$, let $U$ be the $x$-axis, and $W$ the $x,y$-plane. Then $U^\perp$ is the $y,z$-plane, and $W^\perp$ is the $z$-axis.
//Edit: I was slow so I missed Joans Meyer's edit, which kind of makes my answer redundant.
