Is $K\subset\mathbb{R}^2$ homeomorphic to an interval if $K$ is connected but $K\setminus\{x\}$ is not for any $x\in K$? Must it have empty interior?

Given that $K$ is a connected subset of $\mathbb{R}^2$ such that $\forall x\in K, K\setminus\{x\}$ is not connected, then

1. K must be homeomorphic to an interval of $\mathbb{R}$
2. K must have empty interior.

Well, I feel that 1 is correct but I'm not able to make it formal, and I'm not sure about 2.

Thank you for help.

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Is $K$ connected? – Mercy King Jul 8 '12 at 23:09
2. is correct: removing a point of interior does not disconnect a set, because a disk minus a point is still connected. – user31373 Jul 8 '12 at 23:15
@LeonidKovalev You are assuming that $K$ is connected. If a set is not connected, it is possible that removing a point from the interior of the set keeps the set disconnected. – William Jul 8 '12 at 23:17
@LeonidKovalev Nevermind, the post has been edited to say that $K$ is connected. – William Jul 8 '12 at 23:17
1 is false: take the union of two axes. – user31373 Jul 8 '12 at 23:17