If a 3D surface is a line in any 2D-subspaces, then the surface must be a plane. $H$ is a 2D plane.
Let $X\subset \mathbb R^3$ be a connected surface. Given that $X\cap H$ is always a line or a plane or emptyset for any $H$, then what can we conclude for the shape of $X$? 
I think it is very obvious that $X$ must be plane. However I have no idea how to start it.
 A: This is inspired by the comment of @DonThousand although I believe his comment (as written) has a mistake or maybe a typo.

Claim: If for any plane $H, X \cap H$ is empty, a line, or a plane, then $X$ is empty, a plane, or the entire 3D space.  (None of the planes mentioned need to pass through the origin.)

Proof by case analysis:
If $X$ contains zero points, then it is empty.  This case is possible as it satisfies the precondition.
$X$ cannot contain exactly $1$ point: pick any plane $H$ through that point and $X\cap H$ violates the precondition.  
For the rest, assume $X$ contains at least $2$ points $v, w$ where $v \neq w$.  Pick any $H$ through $v, w$.  Since $X \cap H$ is non-empty, by the precondition it must be a line (the line $L$ through $v, w$) or the plane $H$.  In either case we have $L \subset X \cap H \subset X$.
$X$ cannot be just the line $L$: pick any plane $H' \perp L$ and $X \cap H'$ would be a point, violating the precondition.
So $X$ contains a point $u \notin L$.  By an analogous argument to above, $X$ contains every line defined by $\{u, x\}$ for all $x \in L$.  This is almost a plane already, except for the line $K$ through $u$ and $|| L$.  However, Just pick another point $u'$ outside $K$ and $L$ and using $u'$ and $L$ we can "sweep" again and include $K$.  In short, $u \in X, u \notin L \implies X$ contains the plane $Y$ defined by $\{u, L\}$.
If $X = Y$, this is possible as it satisfies the precondition.
If $X$ contains another point $p \notin Y$, then by an analogous argument, $X$ contains the entire 3D space.
Thus: $X$ is emtpy, or a plane, or the entire 3D space.
