How about some tomography? This should work if $X$ is open. Assume $X\subset \mathbb R^3$ is nonempty and for every plane $H$ the intersection $H\cap X$ is either contractible or empty. (Note that an open subset of the plane is contractible if it is ((nonempty,) connected, and) simply connected.)

Claim 1: $X$ is contractible.

Proof: Consider the space of all pairs $(x,H)$, $x\in X$ and $H$ a plane containing $x$.

Fiber this by $(x,H)\mapsto x$. It's a trivial bundle over $X$ with fiber $P^2$.

Fiber the same thing by $(x,H)\mapsto H$. The nonempty fibers are contractible, so the domain is homotopy equivalent to the image, which in turn fibers over $P^2$ with contractible fibers.

So $X\times P^2$ has the same homotopy type as $P^2$, and upon further inspection $X$ is contractible.

Claim 2: $X$ is convex.

Proof: Let $L$ be any line whose intersection with $X$ is nonempty, and play the same game again with pairs $(x,H)$, but now the plane $H$ is constrained to contain $L$. Call the space of all such pairs $Y$. On the one hand, $Y$ is equivalent to the circle $P^1$ by the same kind of argument as before. On the other hand, $Y$ is the blowup of the $3$-manifold $X$ along the $1$-manifold $X\cap L$. The complement $Y'$ of $(X\cap L)\times P^1$ in $Y$ is the same as the complement of $X\cap L$ in $X$, so it is connected. Therefore all of the group $H_1(Y,Y')$ comes from $H_1(Y)$. But the latter group is of rank $1$ while the former is isomorphic to $H_0( (X\cap L)\times P^1)$. (I am using mod $2$ coefficients.) It follows that $X\cap L$ is connected.

Maybe this generalizes to $\mathbb R^n$. The hypothesis I am thinking of is that every nonempty hyperplane section is contractible, or equivalently $(n-2)$-connected. I don't believe that every hyperplane section simply connected is enough if $n>3$

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