Consider a topological space $X$ with the property that each $x\in X$ has a connected neighborhood. Suppose that $X$ is second countable and let $f:X\to\mathbb{R}$ be continuous. Show that $f(X)=U\cup Q$, where $U$ is open and $Q$ is countable.
My attempt:
Note: I have already proved that $X$ has a countable number of connected components and that all components are open. These results can be used for this problem.
Suppose that $\{C_n: n\in\mathbb{N}\}$ is the set of components of $X$. Then, $f(C_n)$ is a connected set in $\mathbb{R}$, and therefore an interval, for all $n\in\mathbb{N}$. Components form a partition of the space, so we can write $X=\bigcup_n C_n$, so that $f(X)=\bigcup_n f(C_n)$. We need a countable set $Q$, and we know that $\mathbb{Q}$ is dense in $\mathbb{R}$, so we could possible do the following:
Let $n\in I_1 \iff f(C_n)$ is an open interval in $\mathbb{R}$, then $\bigcup_{n\in I_1} f(C_n) =: U\in\tau_{\mathbb{R}}$. Let $I_2 = \mathbb{N}\backslash I_1$. Then $\forall n\in I_2: \exists q_n\in f(C_n)^{°}\cap \mathbb{Q}\subset f(C_n).$ So $Q := \bigcup_{n\in I_2} f(C_n)$ will be at most countable.
Is this correct?
No it;'s simpler: $f[C_n]$ is connected, so it is either a singleton, or a non-degenerate interval, which we can always write as $U_n$ (open, possibly empty), unioned with a finite set $F_n$. (e.g. $[a,b]= \{a,b\} \cup (a,b)$ or $[a,+\infty) = \{a\} \cup (a,+\infty)$ etc. $F_n$ can be empty if $f[C_n]$ is already open, of course.)
But then $f[X]=f[\bigcup_n C_n] = \bigcup_n f[C_n]= (\bigcup_n U_n) \cup (\bigcup F_n)$ which is of the form "open union countable".
That the $C_n$ are open in $X$ is itself not needed here, just that they are connected and cover $X$. The fact that they are open only plays a rôle in the lemma that there are only countably many such components (from second countability, or ccc-ness, of $X$).