This will not always work. Take $V=W=\mathbb R^2$, and then something like $v_1=(1,0)$, $v_2=(0,1)$, $v_3=(1,1)$, $v_4=(1,2)$, $v_5=(2,1)$, and let $X$ be the smallest set that contains $(v_j, v_j)$ ($j=1,2,3$), $(v_4,v_5)$ and satisfies your conditions. In other words, the vectors here can be multiplied by arbitrary numbers, but that's it, since condition (3) never applies.

Then for example $(v_4,v_4)\notin X$, but any bilinear form that vanishes on $X$ is identically equal to zero (check that the four vectors above span the tensor product $V\otimes V$ when thought of as elements of this space, or, easier still perhaps, work with the matrix representation of the bilinear form with respect to the standard basis $v_1,v_2$).

This will not always work. Take $V=W=\mathbb R^2$, and then something like $v_1=(1,0)$, $v_2=(0,1)$, $v_3=(1,1)$, $v_4=(1,2)$, $v_5=(2,1)$, and let $X$ be the smallest set that contains $(v_j, v_j)$ ($j=1,2,3$), $(v_4,v_5)$ and satisfies your conditions. In other words, the vectors here can be multiplied by arbitrary numbers, but that's it, since condition (3) never applies.

Then for example $(v_4,v_4)\notin X$, but any bilinear form that vanishes on $X$ is identically equal to zero (check that the four vectors above span the tensor product $V\otimes V$ when thought of as elements of this space, or, easier still perhaps, work with the matrix representation of the bilinear form with respect to the standard basis $v_1,v_2$).

A more highbrow answer is also possible: If $K=\mathbb C$, we can take $$ X = \{ (a\overline{v}, bv): a,b\in\mathbb C, v\in V \} . $$ Here $\overline{v}$ means the complex conjugate of $v$, taken componentwise, with respect to a fixed basis. If $B$ denotes the bilinear form, then this choice of $X$ makes the sesquilinear form $S(v,w)=B(\overline{v},w)$ vanish on all $(v,v)$, so $S$ is identically zero by polarization.

This will not always work. Let $\{ e_1, e_2\}$ be the standard basis of $\mathbb R^2$ and rotate by 45 degrees, say, to obtain the basis $\{ f_1, f_2 \}$. Let $X$ be the smallest set that contains $(e_1,e_1)$, $(e_2,e_2)$, $(f_1,f_2)$, $(f_2,f_1)$ and satisfies your conditions (so we can put arbitrary constants in front of these vectors, but that's it since the third condition never applies). Then for example $(f_1,f_1)\notin X$.

However, it's easy to check that these four vectors are a basis of the tensor product (the first two and the last two are linearly independent, and the corresponding subspaces don't intersect), so every bilinear form that vanishes on $X$ is identically equal to zero.

This will not always work. Take $V=W=\mathbb R^2$, and then something like $v_1=(1,0)$, $v_2=(0,1)$, $v_3=(1,1)$, $v_4=(1,2)$, $v_5=(2,1)$, and let $X$ be the smallest set that contains $(v_j, v_j)$ ($j=1,2,3$), $(v_4,v_5)$ and satisfies your conditions. In other words, the vectors here can be multiplied by arbitrary numbers, but that's it, since condition (3) never applies. 

Then for example $(v_4,v_4)\notin X$, but any bilinear form that vanishes on $X$ is identically equal to zero (check that the four vectors above span the tensor product $V\otimes V$ when thought of as elements of this space, or, easier still perhaps, work with the matrix representation of the bilinear form with respect to the standard basis $v_1,v_2$).