Cardinality of set of all everywhere-discontinuous functions

Your intuition is correct. Here's one way to prove it:

Write $\mathbb{Q}$ as the disjoint union of two dense sets $A, B$ (e.g. take $A$ to be the dyadic rationals and $B=\mathbb{Q}\setminus A$). Then:

Any function $f$ satisfying $f(a)=1$ for $a\in A$, $f(b)=0$ for $b\in B$ is everywhere discontinuous.

So how many functions of this type are there? Well, there's no restriction on the behavior of $f$ on irrational inputs, so we have:

The number of everywhere discontinuous functions is at least the number of functions from the irrationals to the reals.

Now using the fact that the irrationals have cardinality $c$, do you see how to finish the proof?

Given a subset $A$ of $\mathbb{R} \setminus \mathbb{Q}$, I can produce a unique function which is everywhere discontinuous: namely, the function which is $0$ on $\mathbb{Q}$, $1$ on $A$, and $-1$ everywhere else.

Therefore there are at least as many everywhere-discontinuous functions as there are subsets of the continuum-sized $\mathbb{R} \setminus \mathbb{Q}$.

Start with Conway's base 13 function $c $ (whose range on any interval is all of $\mathbb R $), which is everywhere discontinuous, and note that if $f $ only takes values $0$ and $1$, then $c+f $ is again everywhere discontinuous (since its range on any interval is unbounded). Now note that there are $2^\mathfrak c $ such functions $f $: the characteristic functions of subsets of $\mathbb R $. Since this is an upper bound (being the total number of functions from $\mathbb R $ to itself), we are done.