Property similar to connectedness Recall that $X$ is connected if $X$ cannot be written as the union of nonempty open sets with empty intersection.
Consider the following similar property:

$X$ is good if $X$ cannot be written as the union of connected open sets with disconnected intersection.

In other words, $X$ is good if, for all open connected sets $A$ and $B$ with $A\cup B=X$, we have that $A\cap B$ is connected.
$\Bbb R^n$ is good for all $n$. $S^n$ and $\rm\Bbb RP^n$ are good for all $n\ge2$. In fact, if $H_1(X;G)=0$ for some $G$, then $X$ is good. This comes from the portion of the reduced Mayer–Vietoris sequence around $\widetilde H_0(A\cap B)$.
Are these the only good spaces? Or are there good spaces with $H_1(X;G)\ne0$ for all $G$? I have no idea how to approach this..
 A: $\require{AMScd}$I'd rather think about the negation of your first condition. So call a space bad if there exists a decomposition $A \cup B$ into connected open sets with disconnected intersection. Call a space semi-bad if there exists a decomposition into open sets $A \cup B$ such that the induced map on relative homology $\tilde H_0(A \cap B) \to \tilde H_0(A) \oplus \tilde H_0(B)$ is not injective.
As you say in your post, semi-bad implies that $H_1(X;G)$ is nonzero for all $G$. 
Theorem: If $H^1(X;\Bbb Z) \neq 0$, and $X$ is a CW complex, then $X$ is semi-bad.
This is because there are canonical bijections between $$H^1(X;\Bbb Z) \cong \text{Hom}(H_1(X;\Bbb Z),\Bbb Z) \cong [X,S^1].$$
(The first isomorphism is the universal coefficient theorem, the second is an appendix in Hatcher chapter 1 + the fact that $H_1$ is the abelanization of $\pi_1$. The latter bijection sends each homomorphism to a map that induces that homomorphism.)
So our assumption implies that there is a homotopically nontrivial map $p: X \to S^1$. Decompose $S^1$ into $U \cup V$ the obvious slight thickenings of the northern and southern hemispheres. Then $p^{-1}(U) \cup p^{-1}(V)$ is a semi-bad decomposition of $X$. To see this, note first that Mayer-Vietoris is natural under maps of triples $(X,U,V)$, so we get a diagram
$$\begin{CD}
H_1(X;\Bbb Z) @>>> \tilde H_0(p^{-1}(U) \cap p^{-1}(V);\Bbb Z) @>>> \tilde H_0(p^{-1}(U)) \oplus H_0(p^{-1}(V);\Bbb Z)\\
@Vp_*VV @VVV @VVV \\
\Bbb Z =H_1(S^1;\Bbb Z) @>\cong>> \tilde H_0(U \cap V;\Bbb Z) @>>> H_0(U;\Bbb Z) \oplus H_0(V;\Bbb Z) = 0\\
\end{CD}$$
A diagram chase along with the fact that $p_*$ is nonzero easily shows from here that this is a semi-bad decomposition of $X$.

We should now relate this to your condition: that $H_1(X;G) \neq 0$ for all $G$. If $H_1(X;\Bbb Z)$ is finitely generated, this implies that $H^1(X;\Bbb Z)$ is nonzero (invoke the classification theorem of finitely generated abelian groups and your assumption for $G = \Bbb Q$). I have not thought about this in further generality. As a sign of danger, think about the universal coefficient theorem, 3F.12 in Hatcher, and Whitehead's problem, known to be independent of ZFC.
One might also be interested in relating the notion of semi-badness to the notion of badness. I have not made any attempt to do so.
