Connected subspace of a disconnected space My friend and I can't figure out something that seems pretty simple. So any help would be appreciated :)
Does there exist a topological space $X$ and a subspace $A \subseteq X$ such that:


*

*$X$ is connected;

*$A$ is connected;

*$\forall a \in A$ the subspace $A\smallsetminus \{a\}$ is connected and the subspace $X \smallsetminus \{a\}$ is not connected.


Edit:
To make it less trivial, can we find an "interesting" example of a subspace $A$? Suppose one with non-empty interior?
Thanks!
 A: There is at least a trivial sort of example. Let $X$ by the Knaster-Kuratowski fan, and let $A=\{p\}$, where $p$ is the dispersion point $\left\langle\frac12,\frac12\right\rangle$. (I take the empty set to be connected, since it is not the union of two non-empty, pairwise disjoint open sets.)
Added: Here’s a much better example. Let $X=S^1\times\Bbb R$ with the following metric: for $\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\in Y$ let
$$d\big(\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\big)=\begin{cases}
|y_0-y_1|,&\text{if }x_0=x_1\\
|y_0|+|y_1|+\rho(x_0,x_1),&\text{if }x_0\ne x_1\;,
\end{cases}$$
where $\rho$ is any metric on the circle $S^1$ generating the usual topology, e.g., the restriction of the Euclidean metric in the plane. Let $A=S^1\times\{0\}$. $X$, $A$, and $A\setminus\{a\}$ for $a\in A$ are all path-connected and hence connected. However, for $a=\langle x,0\rangle\in A$ the set $X\setminus\{a\}$ has three components, $\{x\}\times(0,\to)$, $\{x\}\times(\leftarrow,0)$, and $(S^1\setminus\{a\})\times\Bbb R$, so it’s not connected.
A: Brian has already shown how this is possible even for infinite subsets $A$ in metric spaces. I'm writing my example, however, which came to my mind before Brian's edit, and which I like due to its simple description.
In the real line identify all points between $0$ and $1$, so $X=\mathbb R/(0,1)$. Let $b$ denote the equivalence class of the interval, and $A=\{0,b\}$. Clearly $X$ is connected. $A$ has the Sierpiński topology, so it is connected, too. But removing any point of $A$ disconnects $X$.
A finite version of this example would be $X=\{-1,0,b,1\}$ with the topology $\tau$ generated by the base $\{\{-1\},\{-1,0,b\},\{b\},\{b,1\}\}$. Let $A=\{0,b\}$. Then every point in $A$ divides $X$ into two components.
