Consider 2 points $x_1$ and $x_2$ on some plane $wx+b = 0$.
If we draw the vector from the origin to each point, the vector is represented by $x_1$ and $x_2$. Isn't this vector on the plane $wx+b=0$? This is my intuition when I draw the vectors.
However, if I subtract $wx_2+b=0$ from $wx_1 + b =0$, I get $$ w(x_1 - x_2) = 0 $$
This plane no longer has the offset term, so it is indeed parallel to $wx+b=0$. But this seems to contradict what I said earlier about how $x_1 - x_2$ being a vector on the original plane?
I think this has to do with the distinction between a point and a vector. If the vector $\vec{x_1}$ points from the origin to a point on the plane and the vector $\vec{x_2}$ points from the origin to another point on the plane. This means the vector $\vec{x_1} - \vec{x_2}$ lies in the plane since it is the vector connecting the two points.
If a vector $\vec{x}$ is a position vector that emanates out of the origin, then the point that $\vec{x}$ points to lies on the plane if it satisfies the equation $\vec{w} \cdot \vec{x} + b = 0$. Thus, $\vec{w} \cdot \vec{x_1} + b = 0$ and $\vec{w} \cdot \vec{x_2} + b = 0$ for position vectors $\vec{x_1}$ and $\vec{x_2}$ pointing to points on the plane.
Which means $\vec{w} \cdot (\vec{x_1} - \vec{x_2}) = 0$.
All this says is that the vector $\vec{x_1} - \vec{x_2}$ is perpendicular to the vector $\vec{w}$ which is the normal vector to the plane. You assumed that $\vec{x_1}$ and $\vec{x_2}$ point to points on the plane. $\vec{x_1} - \vec{x_2}$ being perpendicular to the normal vector $\vec{w}$ does not contradict this.
Actually, $\vec{x_1}$ and $\vec{x_2}$ are position vectors from the origin to points on the plane. So they satisfy a different equation than $\vec{x_1} - \vec{x_2}$, which lies in the plane, does. This doesn't mean $\vec{x_1} - \vec{x_2}$ is not in the same plane that $\vec{x_1}$ and $\vec{x_2}$ point to, just that it satisfies a different equation to lie in the same plane than $\vec{x_1}$ and $\vec{x_2}$ do.