Consider the following property:
$\mathbb R$ is a connected space, but $\mathbb R\setminus \{p\}$ is disconnected for every $p\in \mathbb R$.
$S^1$ is a connected space and if we remove any point, it is still connected. But if we remove two arbitrary points $p$ and $q$, the resulting $S^1 \setminus \{p,q\}$ is disconnected.
Let $X$ be a topological space. Let's call $X$ to be $n$-flimsy if removing fewer then $n$ arbitrary points leaves the space connected and removing any $n$ arbitrary (distinct) points disconnects the space.
We saw that $\mathbb R$ is $1$-flimsy and $S^1$ is $2$-flimsy (as $S^1 \setminus \{*\} \cong \mathbb R$).
Question: Is there a $3$-flimsy space?
So I'm searching for a space $X$ such that the removal of any $3$ points disconnects the space, but fewer don't.
I suspect that there is no such space. I thought I could show it by showing first, that $1$- or $2$-flimsy spaces are in some way unique, but I found many examples of $1$-flimsy spaces which are significantly different (the long line, a variant of the topological sinus, trees).
Alternatively: Is there a standard terminology for this property? (it definitely 'feels' like $n$-connectivity in graph theory)
Addendum 1: A space $X=\{x,y\}$ with two points is a trivial $3$-flimsy example, since we cannot remove three distinct points. Of course, I'm interested in real examples.
Addendum 2: Since Qiaochu Yuan and Paul Frost argued that CW-complexes won't work, here are some thoughts concerning the finite case:
Let $(X,T)$ be a topological space with finite $X$. Then $T$ is automatically an Alexandrov topology and therefore has the Specialization preorder $\prec$. If we have a connected component $Z(x)$ of a point $x$ in a finite space with Alexandrov topology, then $Z(x)$ and its complement are closed and open, so they are downwardly closed. If we visualize $(X,T)$ by the graph $G$ which has $X$ as vertices and two vertices $v,w$ are connected if $v\prec w$ or $w \prec v$, then connected components in $T$ refer to connected components of the graph. Deleting a point in $X$ corresponds to deleting the respective vertex.
Claim: There is no finite $1$-flimsy space (disregarding the trivial examples above). Otherwise we have a graph where the removal of any vertex results in a disconnected graph. This graph can't be finite.
Corollary: There are no finite $n$-flimy spaces for $n\in \mathbb N$ (disregarding the trivial examples above). The removal of one point results in a finite $n-1$-flimsy space, which can't exist (induction).
Still open: Are there nontrivial $3$-flimsy spaces? Those should be infinite and shouldn't be homeomorphic to CW-complexes.
Addendum 3: Funfact: Every topological space can be embedded into a $1$-flimsy space. Just add a real line to each point (as a one-point union). Alternatively, add $1$-spheres to every point. Then add $1$-spheres to each new point. Continue like this for eternity.
Addendum 4: In the setting of Whyburn's book Analytic topology it is shown, that a compact set cannot be $1$-flimsy (Chapter 3, Theorem 6.1). Since all my examples for $1$-flimsy spaces are non-compact: Is there an example of a compact $1$-flimsy space? Are all $n$-flimsy spaces non-compact (at least they are infinite)?