Is this function constructed using AC necessarily discontinuous everywhere? Assume AC.  Let $x_\alpha$ be a well-ordering of $\mathbb{R}$.  For all $\alpha < \mathfrak{c}$, let $F(x_\alpha) = x_{\alpha+1}$.
Can it be proven that $F$ is discontinuous everywhere?
 A: $F$ can be continuous at some points, if the well-ordering is defined in the right way.  For example, choose your well ordering so that each real number in $(0,1)$ is a unique limit ordinal in $c$.  That is, if $T:\mathbb{R} \rightarrow c$ maps each element $x \in \mathbb{R}$ to the element $\alpha$ of $c$ such that $x_\alpha=x$, then we want $x \in (0,1) \Rightarrow \alpha$ is a limit ordinal or 0.  
There are enough limit ordinals to accomplish this, because the cardinality of $c$ is the cardinality of $\mathbb{N} \times L(c)$ where $L(c)$ is (the set of all limit ordinals in $c$) $\cup$ 0.  So the number of limit ordinals must have the same cardinality as $c$, and the same cardinality as (0,1).  
Further define $T$ so that $T(x) = T(x-2)+1$ for all $x \in (2,3)$. At this point $T$ is still injective, because $T(x-2)+1$ is not a limit ordinal.  
Now extend $T$ to the rest of $\mathbb{R}$ where it hasn't already been defined in such a way that it is bijective.  
With $x_\alpha$ defined in terms of this $T$, F will be continuous on $(0,1)$.  In fact, it will be identically equal to $f(x)=x+2$ on $(0,1)$.  
