Show the Set is Connected Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. Show the set
$$A:= \{\alpha \in \mathbb{R}: \exists \{x_n\}_{n \in \mathbb{N}} \subset \mathbb{R} \, \, \text{ with} \, \, \lim_{n \to \infty} f(x_n) = \alpha \}$$  
is connected.  
Proof idea:   
Since this is a subset of $\mathbb{R}$. This is equivalent to showing that, given $\alpha_1, \alpha_2 \in A$, if $\alpha_1 < z < \alpha_2$ then $z \in A$. Let $\epsilon < \min\{d(\alpha_1,z), d(\alpha_2,z)\}$. Let $\{x_n\} \to \alpha_1$ and $\{x'_n\} \to \alpha_2$. There exists an $N \in \mathbb{N}$ so that, for $f((x_N,x'_N)) \subset (\alpha_1 -\epsilon, \alpha_2-\epsilon)$.
Since $f$ is continuous, the IVT gives that there exists $z_1 \in (x_N,x'_N)$ with $f(z_1) = z$. Iteratively, this gives a sequence $\{z_n\}$ with $f(z_n) = z_0$ and $\lim \limits_{n \to \infty} f(z_n) = z$. Ergo $z \in A$ so $A$ is connected.  
Any issues with this proof? Any other ways to attack this problem?

$A$ is the closer of $f(\Bbb R)$  i.e $A=\overline{f(\Bbb R)}$.
As $\Bbb R$ is connected and $f$ continuous, then $f(\Bbb R)$ is connected, hence the closer is connected.
