'flimsy' spaces: removing any $n$ points results in disconnectedness

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.

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.