I would like to know an explicit method on constructing an everywhere discontinuous real function $F$ with the property: $$F((a+b)/2)\leq(F(a)+F(b))/2.$$
There is a non-constructive example (with the inequality been trivial):
Take a Hamel basis $S$ of the $\mathbb{Q}$-linear space $\mathbb{R}$, take in $S$ a countably-infinite subset $X=\{x_1, x_2, ...\}$, then by multiplying a rational number $c_n$ to $x_n$ for each $n$ we can produce a set $Y=\{y_1, y_2, ...\}$ with $0< y_n\leq1/n$. Now we replace the $X$ in $S$ with $Y$ and obtain a new Hamel basis $T$. Take a $t_0\in T$; let $F(t_0)=0$ and let $F(t)=1$ for any other $t\in T$, then $F$ extends to a function on $\mathbb{R}$ linearly, and it is clear this is a required function.
By the answer of Conifold below, such an explicit method does not exist. But it would still be nice to know how to give such a non-constructive function with the inequality been strict.
No constructive example exists. Functions satisfying this inequality are called midpoint convex. Lebesgue measurable functions that are midpoint convex will be convex by a Sierpinski Theorem, therefore not just continuous but differentiable at all but countably many points.
There is a model of set theory, called Solovay model, where all axioms are satisfied except for the axiom of choice, but every function is Lebesgue measurable. In this model only convex functions will be midpoint convex. Therefore existence of everywhere discontinuous midpoint convex functions can only be proved with the axiom of choice, i.e. non-constructively.