Let $\mathbb{R}\mathbb{P}^3$ be the real projective three-space. It is clear that any two hyperplanes in $\mathbb{R}\mathbb{P}^3$ intersect. But I wonder whether one could embed two copies of the real projective plane somehow to $\mathbb{R}\mathbb{P}^3$ so that the images are disjoint:
Can I find two subsets $P_1, P_2 \subseteq \mathbb{R}\mathbb{P}^3$ which are disjoint and both homeomorphic to the real projective plane $\mathbb{R}\mathbb{P}^2$?
If you could embed two copies of $\Bbb{RP}^2$ such that they don't intersect, you could demand that the second one be outside a (sufficiently small) tubular neighborhood of the first. The tubular neighborhood of any embedded $\Bbb{RP}^2$ is nontrivial by Stiefel-Whitney number manipulations. The boundary of such a tubular neighborhood is a 2-sphere. $\Bbb{RP}^3$ is an irreducible 3-manifold, so this 2-sphere splits $\Bbb{RP}^3$ into two sides, one of which is a 3-ball. The second $\Bbb{RP}^3$ sits inside the 3-ball (since it's on the other side of the boundary of the tubular neighborhood than the first $\Bbb{RP}^3$!). This is a contradiction - $\Bbb{RP}^2$ does not embed into $\Bbb R^3$.
This same argument generalizes. A copy of $\Bbb{RP}^n$ ($n>1$) in $\Bbb{RP}^{n+1}$ has a tubular neighborhood; the boundary of the tubular neighborhood is a sphere $S^n$ that separates the manifold. We want to figure out what's on the other side. Lift this sphere to the universal cover; by the Schoenflies theorem it bounds a (topological, at least) ball. You can verify that the restriction of the covering map to this ball is still a covering map, and hence one-sheeted, so that we had bounded a (topological) $(n+1)$-ball in $\Bbb{RP}^{n+1}$ as well. Now $\Bbb{RP}^n$ can't embed into $\Bbb R^{n+1}$, even topologically. So you cannot embed two disjoint copies of $\Bbb{RP}^n$ inside $\Bbb{RP}^{n+1}$, $n>1$.
(You can obviously embed a bunch of disjoint circles in $\Bbb{RP}^2$. However, in the spirit of the question, if you asked "can I disjointly embed two copies of $\Bbb{RP}^1$ that have the Mobius band as their normal bundle?", the answer would be no, by the same argument + any curve in the plane has trivial normal bundle.)