It's a trivial though verbose to note, that in $\mathbb{R}^2$ if you have a point $p$ contained between parallel lines $l_1, l_2$ and if $p$ is formed by the intersection of 2 rays $r_1, r_2$ (originating from $p$) that at least one of $r_1,r_2$ doesn't intersect both $l_1, l_2$.
In $\mathbb{R}^3$ a similar conjecture can be made: if you have a point $p$ contained between 2 parallel planes $l_1, l_2$, and $p$ is the unique extreme point formed by the intersection of the 3 half-spaces $h_1, h_2, h_3$, whose boundary planes $\partial h_1, \partial h_2, \partial h_3$ are in general position, call their intersection $c$ (for cone). Then there exists a half-space, $h_i, i \in {1,2,3}$ such that $\partial h_i \cap c$ doesn't intersect both planes $l_1, l_2$.
Though I haven't made it rigorous, this question in $\mathbb{R}^3$ seems to reduce to the following idea:
In $\mathbb{R}^2$ it is impossible to cut a circle $C$ with center $o$ into 3 non-zero radial sectors such that if you pass 2 parallel lines $l_1, l_2$ through circle (and let $o$ be located between $l_1, l_2$), then one of the sectors doesn't intersect both lines.
How do I take this rather intuitively obvious fact and prove it? One way is to connect the formal question I gave to the "reduced idea" involving circles, but while the bridge "feels" obvious (its just flattening the 3d problem), sadly it is not clear to me how to rigorously reduce it.
Your statement is false (Try to see why the argument extends to higher dimensions.)
Pick the point to be the origin $O$ and the two parallel planes to be right above and right below the point. Pick three lines that intersect $O$ and are linearly independent. (E.g. one with vector pointing along $(0,1,1), (1,0,1), (0,0,1)$.) Now, you can extend these lines to be half-planes in such a way that $O$ is their intersection. (How?) However, each of these lines intersect both of the parallel planes and hence the half-planes intersect the parallel planes.