Prove that $X$ doesn't have cut points

Solution 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.