Removal of an arbitrary point of the boundary of a closed and connected $A\subseteq\Bbb R^2$ so the new set remains connected

Prove the following statement or find a counterexample:

Let $A\subseteq\Bbb R^2$ be a closed and connected set. Then,$\exists c\in\partial A$ s. t. $A\setminus\{c\}$ is still connected.

I think I found some counterexamples:

$x$ or $y$ axis or any other line in $\Bbb R^2,$ as well as graphs of unbounded continuous functions defined on an open interval $I\subseteq\Bbb R$ or graphs of continuous functions defined on the whole $\Bbb R.$

Question :

Is the unboundedness necessary for the statement not to hold?

I found the following answer in Whyburn's book Analytic topology in Chapter 3, Theorem 6.1 :

Every continuum has at least two non-cut points.

Where, (Chapter 1, 10.)

A compact connected set will be called a continuum

and, (Chapter 3, 1.)

If $M$ is a connected set and $p$ is a point of $M$ such that the set $M \setminus p$ is not connected, then $p$ will be called a cut point of $M$.

Now, if we take $A \subseteq \mathbb{R}^2$ closed, connected and bounded, $A$ is compact by the Heine-Borel theorem. So it is a continuum and it has two non-cut points which means that there exists (at least two) points $c$ such that $A \setminus c$ is connected. This solves the case $\mathring{A} = \emptyset$ which intuitively corresponds to curves in $\mathbb{R}^2$.

For the general case, I think that the theorem still holds because $\mathring{A} \neq \emptyset$ looks like a strong condition to me but I haven't been through yet.