'flimsy' spaces: removing any $n$ points results in disconnectedness 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)?
 A: If I did not make any mistake, 3-flimsy spaces does not exist. You can check this link for my proof and some other results about 2-flimsy spaces. Without giving all the details, here are the big steps of the proof:
First, we show that if $X$ is a 2-flimsy space and $x\neq y\in X$, then $X\backslash\{x,y\}$ has exactly two connected components. For this, we consider 3 open sets $U_1,U_2,U_3$ such that $(U_1\cup U_2\cup U_3)\cap\{x,y\}^{c}=X\backslash\{x,y\}$, $U_1\cap U_2\cap\{x,y\}^{c}=U_1\cap U_3\cap\{x,y\}^{c}=U_2\cap U_3\cap\{x,y\}^{c}=\emptyset$, and $\forall i\in\{1,2,3\},\ U_i\cap\{x,y\}^{c}\neq\emptyset$. If $u_1\in U_1\cap\{x,y\}^{c}$ and $u_2\in U_2\cap\{x,y\}^{c}$, then we can show $X\backslash\{u_1,u_2\}$ is connected.
The second big step is to consider $x,t,s\in X$, three distinct points of a $2$-flimsy space. We denote $C_1(t),C_2(t)$ the two connected components of $X\backslash\{x,t\}$ and $C_1(s),C_2(s)$ the two connected components of $X\backslash\{x,s\}$. We suppose $s\in C_1(t)$ and $t\in C_1(s)$. Then $D=C_1(t)\cap C_1(s)$ is one of the two connected components of $X\backslash\{t,s\}$. In fact, the finite number of connected components implies $C_2(t)\cup\{x\}$ is connected, so the same goes for $(C_2(t)\cup\{x\})\cup(C_2(s)\cup\{x\})$ : the only thing to verify is the connectedness of $D$. The proof looks like to the first step. If $U,V$ are two open sets of $X$ such that $U\cap V\cap D=\emptyset$, $(U\cup V)\cap D=D$, and $U\cap D\neq\emptyset$ and $V\cap D\neq\emptyset$, and if $u\in U\cap D$ and $v\in V\cap D$, then we show $X\backslash\{u\}$ or $X\backslash\{v\}$ is not connected.
Finally, if $X$ is a $3$-flimsy space and $x,y,t,s$ some distinct points of $X$, then $D$ (defined as previously in $X\backslash\{y\}$, a 2-flimsy space) is open and closed in $X\backslash\{x,t,s\}$ and in $X\backslash\{y,t,s\}$, so it is open and closed in $X\backslash\{t,s\}$, which is not connected. So $X$ is not a 3-flimsy space after all.
A: Here is a proposition that I believe will help to at least figure out whether or not a $3$-path-flimsy space exists.
An $n$-path-flimsy space would be a space such that removing fewer than $n$ points would keep the space path-connected, but removing any $n$ points would make the space not path-connected.
Proposition A: Let $X$ be a $2$-path-flimsy space and $x\in X$.
Then for any path-connected open neighborhood $N$ of $x$, such that $X\setminus N$ is also path-connected, the space $N\setminus\{x\}$ has at most two path-connected components.
Proof of Proposition A: The proof is by contradiction.
Assume for the contrary that there exists $x\in X$ with a path-connected open neighborhood $N$, such that $X\setminus N$ is also path-connected, and such that the space $N\setminus\{x\}$ has three distinct path-connected components $C_1$, $C_2$ and $C_3$.
Let $c_i\in C_i$.
Since $X$ is $2$-path-flimsy, the space $X\setminus\{x\}$ is path-connected, so $N\neq X$, so we can find $p\in X\setminus N$.
Fix some $1\leq i\leq3$.
Since $N$ is path-connected, it follows that the set $C_i\cup\{x\}$ is path-connected.
This is because there is a path from $x$ to $c_i$ in $N$, and we can deduce that the last moment the path was not in $C_i$, it must have been at $x$ by the definition of path-connected component.
By similar reasoning, we find that $C_i\cup(X\setminus N)$ is path-connected.
Since $C_i$ is a path-connected component of $N\setminus\{x\}$, any path that leaves $C_i$ must pass through $X\setminus(N\setminus\{x\})=\{x\}\cup(X\setminus N)$ first.
Since $X$ is $2$-path-flimsy, $X\setminus\{c_i\}$ is path-connected, so from any $c\in C_i$ there is a path that leaves $C_i$.
We can conclude that there is either a path from $c$ to $x$ in $C_i\cup\{x\}$, or there is a path from $c$ to $p$ in $C_i\cup(X\setminus N)$.
We can now conclude that $X\setminus\{c_1,c_2\}$ is path-connected, which contradicts the fact that $X$ is $2$-path-flimsy and finishes the proof of Proposition A.
This is because every point is either path-connected to $x$ or $p$ and $c_3$ is path-connected to both.
