If the potential is continuous across a surface, why is the electric field discontinuous then?

Purely mathematically, consider the function $$f(x) = frac{x + |x|}2 = begin{cases}0, quad xle0 x, quad x > 0end{cases}$$ That function is continuous, but its derivative is not: $$frac{df(x)}{dx} = begin{cases}0, quad x<0 1, quad x > 0end{cases}$$ In fact, its derivative isn't even defined at the point $x=0$ (note that "$le$" from the first equation got replaced with "$<$" in the second).

Now, back to physics, consider the potential: $$V(mathbf r) = begin{cases}0, quad r_x le0 -E r_x, quad r_x > 0end{cases}$$ The potential is continuous and is something that describes the situation involving a conductor. However, the electric field is: $$mathbf E(mathbf r) = begin{cases}0, quad r_x < 0 E hat{mathbf x}, quad r_x > 0end{cases}$$ In fact, analogous to the purely mathematical example, electric field is not only discontinuous, but it also isn't even defined on the surface $r_x = 0$ (again, "$le$" was replaced with "$<$"), but that is not a problem since a surface is something without any thickness.

The point is that $V_{text{above}} = V_{text{below}}$ doesn't imply $-nabla V_{text{above}} = -nabla V_{text{below}}$ and in fact it can't imply that when $nabla V$ isn't even defined, which is the case when $V$ isn't smooth.

One could also consider that in such situations there will be a very thin region in which the potential smoothly changes its derivative and electric field continuously goes from the "above" value to the "below" value. However, even without such considerations everything works out because, once again, a surface is something without any thickness and including such region won't make a relevant change of results in practically any relevant calculations.