Is there a bounded connected set $X$ such that for all point $b$ there exists $r > 0$ such that $X \setminus O(b, r)$ is disconnect？ I find a question by myself, and I do not know if it is an interesting question.
Let $X \subseteq \mathbb{R}^n$ be a bounded connected set. And I define a "bad point" $b \in \mathbb{R}^n$ with respect to $X$ if there exists $r > 0$ such that $X \setminus O(b, r)$ is not connect, where $O(b, r)$ is a ball with center point $b$ and radius $r$. Let $B(X)$ be a set of all "bad point" with respect to $X$.
Strong question: Is there a bounded connected set $X$ such that $B(X) = \mathbb{R}^n$?
Weak question: Is there a bounded connected set $X$ such that $X \subseteq B(X)$?
I have no idea how to solve my question. I give some examples of my definition.
(1) If $X$ is a ball or spherical surface, then $B(X) = \varnothing$.
(2) If $X = \{(x,0, \ldots, 0) \mid x \in [0,1] \}$ is a close line segment, then $$B(X) = \{(x,a_{1}, \ldots, a_{n - 1}) \mid x \in (0,1), a_{i} \in \mathbb{R} \}$$
(3) If $X = \{(x,0, \ldots, 0) \mid x \in (0,1) \}$ is a open line segment, then $$B(X) = \{(x,a_{1}, \ldots, a_{n - 1}) \mid x \in (0,1), a_{i} \in \mathbb{R} \}$$
 A: As commented, the weak question is already solved. 
For the strong one, I think there is a simple example if I've understood your question well. Consider a cross $C$ in $\Bbb{R}^2$ center at the origin, i.e the union of $I_x=\{(x,0)\mid x\in[-1,1]\}$ and $I_y=\{(0,y)\mid y\in[-1,1]\}$.
Any ball that covers $(0,0)$ by not the whole cross would disconnect $C$. But you can see that for any point $b\in\Bbb{R}^2\setminus\{(0,0)\}$, the (closed) ball $O(b,|b|)$ contains $(0,0)$ but not the whole cross (if you mean open ball then just take a slightly bigger radius). Note that for $b=(0,0)$ you can take a small radius and it will work as well. 
So, for every $b\in\Bbb{R}^2$ there exists $r>0$ such that $C\setminus O(b,r)$ is disconnected. This example can be easily generalized to any higher $\Bbb{R}^n$.
A: Your weak question is most assuredly true -- just take $X=\{(x,0,\ldots,0):x\in(-1,1)\}$. Since deleting any sufficiently small neighborhood of any point in $X$ will leave behind two disjoint line segments, we are done.
Your strong questions is interesting, but I don't know how to answer it well, so bear with me as I describe a construction in $\mathbb{R}^2$:
Consider the curve (in polar coordinates)
$$
(r,\theta) = \left(\exp(f(\theta)),\theta\right)
$$ 
where $f(\theta)$ is any continuous, monotonically increasing function with range $\left(-\ln(2),0\right)$, such as an appropriately scaled and translated $\arctan$ function. Take this curve, a circle with radius $1/2$ centered at the origin, and the unit circle, and define $X$ to be their union. 
Where are the "bad points"? If you take a point outside the unit circle, say $(0,2)$, then we can find a radius $r=1+\epsilon$ which will "cut" the spiral off from the unit circle, leaving us a disconnected set. The same will be true for a point inside the circle of radius $1/2$, and a point inside the annulus can have a radius chosen to "cut" the two circular boundaries off from one another. Thus, all of $\mathbb{R}^2$ is "bad".
How do we take this construction and make it work in higher dimensions? There may or may not be an elegant equation for it, but the idea definetely generalizes -- take some increasingly fine "spiral" along the surface of hyper-sphere shells, and then take many such shells nested together, limiting towards to bounding shells, and link them together with line segments.
I wish I could describe that better, but I don't know how. Hopefully your mental picture is the same as mine.
EDIT: Please just use Javi's answer if you want a good example. This is far too complicated, even in $\mathbb{R}^2$.
