Is there an everywhere discontinuous increasing function?

There is no such function. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is strictly increasing. For each $a\in\mathbb{R}$ let $f^-(a) =$ $\lim\limits_{x\to a^-}f(x)$ and $f^+(a) = \lim\limits_{x\to a^+}f(x)$. Then $f$ is discontinuous at $a$ if and only if $f^-(a) < f^+(a)$. Let $D = \{a\in\mathbb{R}:f\text{ is not continuous at }a\}$, and for each $a\in D$ let $q_a$ be a rational number in the non-empty open interval $I_a = (f^-(a),f^+(a))$.

It’s not hard to check that if $a,b \in D$ with $a<b$, then $f^+(a) \le f^-(b)$, so the intervals $I_a$ are pairwise disjoint. This implies that the rational numbers $q_a$ are all distinct. (If you want to be fancy, the function from $D$ to $\mathbb{Q}$ that sends $a$ to $q_a$ is injective.) But there are only countably many rational numbers, so the set $D$ must be countable. In other words, the function $f$ can have at most countably many points of discontinuity. And of course $\mathbb{R}$ is uncountable, so $f$ cannot be discontinuous everywhere.

Another way to see that it is impossible to have uncountably many of these "gaps" is to first consider the restriction of $f$ to a bounded interval $(a,b)$, where it is bounded. If $\varepsilon>0$, the number of points in $(a,b)$ where there is a gap of size greater than $\varepsilon$ is finite, less than $\frac{1}{\varepsilon}(f(b)-f(a))$. Taking countably many $\varepsilon$s going to zero allows you to conclude that there are only countably many discontinuities in $(a,b)$. Taking countably many $(a,b)$s whose union is $\mathbb R$ allows you to finish.