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Continuity and the Axiom of Choice

I have proved a small generalization of Brian's argument, that is, "If $f:X\rightarrow Y$ is sequentially continuous on $X$ and $X$ is separable, then $f$ is continuous on $X$".

Next, I have proved that "If $f:C\rightarrow Y$ is sequentially continuous on $C$ and $C$ is a connected set in $\mathbb{R}$, then $f$ is continuous on $C$". Now, i want to generalize this.

Is every connected set in a separable metric space is separable? (in ZF)

Edit: I don't know if this helps, but actually the statement i want to prove is exactly the same as proving 'Every connected set in a separable complete metric space is separable'.

(It can be proven that 'Every separable metric space has a separable completion' in ZF. In fact, if $\varphi : X \rightarrow X^*$ is an isometry and $X^*$ is a completion of $X$ and $D$ is a countable dense subset of $X$, then $\varphi[D]$ is dense in $X^*$. Since $\varphi$ is an isometry, it maps connected subset to connected subset.)

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Am I to assume you are working without Choice? – inactive... for now Nov 7 '12 at 9:06
@Arthur Yes. Just edited my post. – Katlus Nov 7 '12 at 9:07
Katlus, do the comment on Brian's deleted answer, $\mathbb{R\setminus Q}$ is always separable (consider algebraic numbers, or $\mathbb Q+\pi$), but it is consistent to have a set of real numbers which is inseparable. – Asaf Karagila Nov 7 '12 at 12:32
@Asaf: What would this comment on the deleted answer say? As a sub-10Ker, I (alas) do not have access to these words of wisdom. (Also, is Katlus able to see the comment on the deleted answer?) – inactive... for now Nov 7 '12 at 12:42
@Arthur: Brian wrote (wrongly) that serparability is hereditary [true only with countable choice] and Katlus replied "isn't it consistent that $\mathbb{R\setminus Q}$ is not separable in ZF?". – Asaf Karagila Nov 7 '12 at 12:46

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