Suppose $U \subseteq \mathbb R^n$ ($n \ge 3$) is an open, contractible set. I am thinking about what would be a minimal requirement on dimension for an affine subspace $\mathcal A$ to divide $U$ into disconnected components.
Intuitively, I am thinking in $\mathbb R^3$ an 1-dim affine space should not be able to do so and clearly a hyperplane should be able to divide the set into at least two connected components. But in general, is there a minimal requirement on the dimension?
It takes dimension $n-1$ to do it. If we're cutting out a space of dimension $n-2$, we can rotate around that, and the space stays connected. The argument below shows how.
We can glue together $U$ from various balls $B_i$; if each $B_i\setminus A$ is connected, $U$ is. They're obviously connected when $A$ doesn't intersect that ball, which leaves the case when $A$ and $B_i$ intersect. Transforming this case with a homeomorphism, we can assume WLOG that $B$ is the unit ball centered at the origin and $A=\{(x_1,x_2,\dots,x_n): x_1=x_2=0\}$.
So, now, we construct a path between any two points $(x_1,x_2,\dots,x_n)$ and $(y_1,y_2,\dots,y_n)$ in $B\setminus A$.
First, move from $(x_1,x_2,x_3,\dots,x_n)$ to $(x_1,x_2,0,\dots,0)$ in a straight line. The first two coordinates are unchanged throughout, so we stay away from $A$, and we stay inside $B$ because $B$ is convex.
Second, move inward or outward in a straight line to $\left(\sqrt{\frac{y_1^2+y_2^2}{x_1^2+x_2^2}}x_1,\sqrt{\frac{y_1^2+y_2^2}{x_1^2+x_2^2}}x_2,0,\dots,0\right)$.
Third, rotate around to $(y_1,y_2,0,\dots,0)$, along the path $\sqrt{y_1^2+y_2^2}(\cos\theta,\sin\theta,0,\dots,0)$. This piece of the path is why we needed two degrees of freedom.
Fourth, move from $(y_1,y_2,0,\dots,0)$ to $(y_1,y_2,\dots,y_n)$ in a straight line. The path is complete, and $B\setminus A$ is connected.
If $A$ has dimension $n-1$ instead, it disconnects any open sets it passes through. Consider the case of the unit ball $B$ minus the space $A=\{(x_1,x_2,\dots,x_n): x_1=0\}$. This splits $B$ into two sets, one with $x_1 > 0$ and one with $x_1 < 0$.