Answer
see the details.
Work Step by Step
Since for any $w\in W$, we have
$$\langle 0,w\rangle =0$$
then $0\in W^\perp$ and $W$ in nonempty set.
Now, let $v_1,v_2\in W^\perp$, then for any $w\in W$, we get
\begin{aligned}
\langle v_1+v_2, w\rangle &=\langle v_1,w\rangle +\langle v_2,w\rangle=0
\end{aligned}
hence, $v_1+v_2\in W^\perp$ which means that $W^\perp$ is closed under addition.
For any $v\in W^\perp$, $w\in W$ and $k\in R$, we have
$$\langle kv, w\rangle =k\langle v, w\rangle=0.$$
Hence, $W^\perp$ is closed scalar multiplication. Consequently, $W^\perp$ is a subspace of $W$.
