$H(x,y)$ is a bilinear form.
I have tried to do something similar to standard inner product in $\Bbb{R}$ such as Gram-Schmidt process, orthogonal complement...But they don't work since we don't know whether $H(x,x)>0$ in some steps of them.
What can I do? Could you please give some hints to me? Thank you.
On
First, note that for any $v\in V$, $\left \langle v,\cdot \right \rangle$ is a linear transformation.
Now if $\{w_1,...,w_k\}$ is a basis for $W$, consider the linear transformation $T:V\to \mathbb{R}^k$ given by $$Tv=(\left \langle w_1,v \right \rangle,...,\left \langle w_k,v \right \rangle).$$
The dimension of the image (the rank of $T$) is at most $k$, so by the Rank-Nullity Theorem, the dimension of the kernel (the nullspace of $T$) is at least $1$, i.e. there is a nonzero vector $v$ such that $Tv=0$. But this means $$\left \langle w_i,v \right \rangle=0$$
for each basis element $w_i\in W$ which is sufficient to say that $$\left \langle w,v \right \rangle=0 \hspace{1em} \forall w\in W.$$
A bilinear form $b: V\times V\to \mathbb{R}$ is the same as a linear map $B:V\to V^*$ (define $B(x)(y) = b(y,x)$ to satisfy your condition).
Let $U\subset V^*$ be the subspace consisting in the $\varphi\in V^*$ such that $\varphi_{|W}= 0$. Since $W$ is a proper subspace of $V$, $U$ is a non-zero subspace of $V^*$. That implies that $B^{-1}(U)\subset V$ has dimension at least $1$ : pick a non-zero $v\in B^{-1}(U)$ ; by definition, $b(w,v)=0$ for all $w\in W$.
Note that this doesn't use that $b$ is symmetric or non-degenerate.