An affine hyperplane $A$ in $\Bbb R^n$ can be given by an equation $u^Tx=c$ where $u$ is a normal vector for $A$ and $c\in\Bbb R$.
Consequently, an intersection of affine hyperplanes can be given by the system of affine equations
$$u_1^Tx=c_1\\ \ \ \ \dots \\u_k^Tx=c_k$$
which can be rewritten in matrix form:
$$Ux=c$$
This shows 1)$\iff$4), moreover considering the affine map $\phi:\Bbb R^k\to\Bbb R^n\ \ x\mapsto Ux-c$, this also shows equivalence with 3).
We can easily verify that any of the properties 1)-4) ensure that the set in question is closed under affine combinations.
Conversely, if a nonempty $A$ is closed under affine combinations, pick any $a_0\in A$, then $A-a_0$ is a linear subspace, then e.g. choosing a basis for $A-a_0$ we can get a linear map $\psi_0:\Bbb R^m\to\Bbb R^n$ such that $\mathrm{im}(\psi_0)=A-a_0$, hence the image of the affine map $\psi:=\psi_0+a_0$ is just $A$. This connects to 2).
Finally, assuming 2), $A=\mathrm{im}(\psi)$, then again, $B:=A-\psi(0)$ is a linear subspace, so (by e.g. extending a basis of $B$) we can define a linear map $\phi_0$ which vanishes exactly on $B$, and then the affine map $\phi:=\phi_0+\psi(0)$ will vanish exactly on $A$, proving 3).
Best Answer
Hint: Let $A_1 = x_1 + U_1$, $A_2 = x_2 + U_2$ your two affine subspaces, if $A_1 \cap A_2 = \emptyset$, we are done, otherwise there is an $x \in A_1 \cap A_2$. But then $A_1 = x+ U_1$ and $A_2 = x+ U_2$ ... does this help?