$X \setminus point$ not path-connected implies $X$ simply connected

Question

Per $X$ be a path-connected space. Does $X \setminus point$ not path-connected implies $X$ simply connected? Thinking about curves and lines seem to suggest the truth of the statement, but I think that one can "build" a topology ad hoc to find a counterexample.

Answer 1

It appears that you are asking the following question: Let $X$ be a path-connected topological space such that for every point $x\in X$, $X-\{x\}$ is not path-connected. Is it true that $\pi_1(X)=\{1\}$?

This question has negative answer. Consider $X$ which is obtained from the circle $S^1$ by attaching to every $s\in S^1$ a copy of $[0,1)$, where we identify $0$ with $s$. (Formally speaking, to define this space you start with a space $Y$ which is the disjoint union of $S^1$ and $$\coprod_{s\in S^1} \{s\}\times [0,1)$$ Then introduce the equivalence relation $s\sim \{s\}\times 0$ for each $s\in S^1$. Then $X$ is obtained by equipping $Y/\sim$ with the quotient topology.)

Now, I will leave it to you to check that every point $x\in X$ disconnects $X$ and that there exists a retraction $X\to S^1$. It follows that $X$ is not simply-connected.