Prove that $X$ doesn't have cut points

Question

Let $X$ be a metric continuum such that for every two points $a,b \in X$ the set $X\setminus \{a,b \}$ isn't connected. Prove that $X$ doesn't have cut points.

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First I tried to prove it by contradiction by it didn't work.

Then I thought it more geometrically and tried to show that $X\cong S^1$, but I didn't know how to start, thought somehow to use (Moore's) theorem about the existence of non-cut points or show something with Peano's continuum.

Nothing really worked, I feel like I am missing something obvious here. Any hints on how to begin ?

Answer 1

Fact (a fundamental theorem on continua)

If $X$ is a metric continuum it has a point $p \in X$ so that $p$ is a non-cutpoint (so $X\setminus\{p\}$ is connected).

Suppose now that $X$ (as you assume) has a cutpoint $q$. Suppose $X\setminus\{q\}=U \cup V$ is a disconnection.

$X_1:=U \cup \{q\}$ and $X_2=V \cup \{q\}$ are continua. So by the fact each has a non-cutpoint $x_1 \in X_1, x_2 \in X_2$, say (clearly both different from $q$).

Now $X\setminus\{x_1,x_2\} = (X_1 \setminus \{x_1\}) \cup (X_2 \setminus \{x_2\})$ is the intersecting (in $q$) union of two connected spaces, hence connected, but this contradicts the assumption on $X$.

QED.