Q:

Does cutting an open space not increase the number of holes

Say you have some open connected subset $B \subset \mathbb{R}^n$ and you "cut" it by some open half space $C$ to produce two new open spaces $B_1$ and $B_2$. "Cutting" means to take the intersection of $B$ and $C$ to produce $B_1$ and to take the intersection of $B$ and the interior of the compliment of $C$ to produce $B_2$. More formally:
$$B_1 = B \cap C$$
$$B_2 = B \cap \bar{\,C\,}^{\circ}$$
(I had to invent this definition; if there is an agreed upon name/definition for this please tell me)
Is it the case that that the total number of holes in $B_1$ and $B_2$ is less than or equal to the number of holes in $B$?
This feels intuitively true to me. Slicing a nice loaf of bread in two eliminates holes in the two spaces. My slightly more formal thinking is that if is a hole in $B$ then it has some set of non-contractible paths around it. If all of those paths are cut in half (the paths go along points that do not lie exclusively in either $B_1$ or $B_2$) then a hole is eliminated from $B_1$ and $B_2$ so the total number of holes decreases. If the cut leaves one contractible path (there's a path that lies in either $B_1$ or $B_2$ around that hole in $B$ then that hole is left in one of the two so the number at the very least does not decrease.

A:

You should read about the Mayer-Vietoris sequence for homology groups. Using it one can show that
$$
b_{n-1}(B)\ge b_{n-1}(B_1) + b_{n-1}(B_2),
$$
where $b_i$'s are the $i$-th Betti numbers. If $C$ is bounded by the hyperplane $H\subset {\mathbb R}^n$ then the tail of the MV sequence (over ${\mathbb R}$) becomes
$$
0=H_{n-1}(B\cap H) \to H_{n-1}(B_1) \oplus H_{n-1}(B_1) \to H_{n-1}(B) \to ... 
$$
This implies the inequality above. 
Note that, intuitively speaking, $b_{n-1}$ counts the number of holes in an open subset of ${\mathbb R}^n$. 
On the other hand, if you are interested in $b_1$ (per your comment), then one can construct examples where $b_1(B)=0$ (actually, $B$ is topologically an open ball) while $b_1(B_1)>0, b_1(B_2)>0$, already in the case $n=3$. 
Edit. To get a specific example, take the graph $y=\cos(x), -2\pi\le x\le 2\pi$ and rotate it about the $y$-axis in the 3-space. Thicken the resulting surface. It is your $B$. It looks like the baking dish below (minus the two small handles) when viewed up side down. Now, cut it via the $xz$-plane. (On the picture this will be a horizontal plane cutting the dish in the middle.) Then $b_1(B)=0$, $b_1(B_1)=b_1(B_2)=1$.  (Note that one of the $B_i$'s will be disconnected.)  

