Suppose $(E, B, \pi)$ is a vector bundle of dimension $n$, $\pi: E\rightarrow B$. If $F\subset E$ and for every $x\in B$, the fibre $F_x(:=\pi^{-1}(x)\cap F)$ is a linear subsbace of $E_x(:=\pi^{-1}(x))$, give a example to show that $F$ is not necessarily a sub-bundle of $E$.
$F$ is a sub-bundle of $E$: for any set $U$ in the open covering of $B$, there is a trivialization $h_U:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ such that (k is a fixed natural number) $$ h_U(\pi^{-1}(U)\cap F)=U\times(\mathbb{R}^k\times\{0\}). $$
I am just learning some concept of vector bundle. I have no idea how to construct such an example, and another exercise in my textbook seems to imply such an example does not exsit when $B=\mathbb{R}$. Since we can alwars view a linear space of dimension $n$ as $\mathbb{R}^n$, the main point is to find appropriate $B$.
Appreciate any help or hint. Thanks in advance!