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.
Best Answer
As you were told in the comments, if the point $p$ that you remove from $f\left(S^1\right)$ is an endpoint, your argument fails. Of course, you can say that $f\left(S^1\right)$ is an infinite set, and that therefore you can always remove a point which is not an endpoint. I think that this is the simplest and shortest proof.