Closure of a connected subset of $\mathbb{R}$ is connected? Prove or disprove: The closure of a connected set in $\mathbb{R}$ is always connected.

Response: I don't really have a grasp on this conceptually, but here's an attempt.
Proof: Let $X$ be a connected subset of $R$, and let $Z$ be the closure of $X$. Suppose (to get a contradiction) that $Z$ is disconnected. That means there are two nonempty open subsets of $R$ covering $Z$, call them $U$ and $V$, such that $U\cap Z$ and $V\cap Z$ are nonempty while $U\cap V\cap Z$ is empty.

Then $U$ and $V$ cover $X$ also. And since $U\cap V\cap X \subseteq U\cap V\cap Z$, then $U\cap V\cap X$ is empty. Pick points $u\in U\cap Z$ and $v\in V\cap Z$ (which exist since those sets are nonempty). That means $u$ is in the closure of $X$ and $U$ is an open neighborhood of $u$, so $U\cap X$ is nonempty. Similarly, $V\cap X$ is nonempty.

But then the previous paragraph shows that $X$ is disconnected, which is a contradiction, so $Z$ must have actually been connected.
Do you know any definition or property of connectivity? A good one is that if $X$ is connected then any continuous $f : X \to \{0, 1\}$ must be constant. Say $\forall x \in X$, $f(x) = 1$. How can you extend this to the closure of $X$ without destroying continuity? Note that this statement is not restricted to $\mathbb{R}$.
You can actually show that a connected subset in R must be singleton or an interval. In both the cases, closure is also connected.