In a superconductor, $U(1)$ gauge symmetry is spontaneously broken. But $U(1)$ gauge symmetry is responsible for conservation of electric charge. Then it appears to me that the electric charge conservation will be violated which is surely wrong. But I cannot understand how the electric charge remain conserved?
Weinberg says here that
The conservation of a current is usually a symptom of some symmetry of the underlying theory, and holds whether or not the symmetry is spontaneously broken.
Let me explain @ACuriousMind 's answer with some verbiage. The short, regrettably oracular, answer is that the Fabri-Picasso theorem does not hold in a finite superconductor, since translational invariance fails at its boundaries. Really, I do appreciate this is aggressively obscure: will strive to explain.
First of all, if you have a chunk of warm superconducting material within an "observation sphere", you could assess the charge of the sphere's contents by Gauss's law. Lowering the material's temperature to a superconductor will not magically create or annihilate charge inside your sphere because somehow U(1) was broken, as you anticipated. In fact, there are (conserved) electric currents that can flow through the superconductor, or course. That's why we love them! You don't even need to measure them directly, as it is precisely such cost-free current loops that arise to oppose and completely cancel an external magnetic field, by Lenz's law, and so exclude it from the bulk of the material, at least beyond a penetration length $\xi$.
So what about the "breaking" of the symmetry? (This is the best argument against using this term, to start with, instead of "hiding" or something less inflammatory.) The point is, as you might see in the WP article cited above, that the symmetry is still there, it's certainly not gone, and its current is conserved! It is just that the symmetry is realized in the nonlinear Nambu-Goldstone mode ($\theta$) which manifests itself somewhat unconventionally, if not obscurely, $\delta \theta\propto \epsilon$.
Something terrible happens when one takes the space integral of the zeroth component of the current to yield infrared divergent answers, even though you could rehabilitate these answers by sticking them in a commutator with fields, in which case they produce the nonlinear transformations of the fields mentioned. A brief proof of this pathology is the F-P argument of the Zen gambit above, but it requires translational invariance, luckily broken (explicitly) by the boundaries of the finite superconductor! The takeaway is that one may, humbly and carefully, talk about conserved charges and certainly conserved currents.
In a superconductor, you have in addition what we now call the Higgs mechanism. (This gem of an article by the late T Kibble, explains the above difficulties in defining Q in infinite, relativistic theories, which are not our concern here.) The London brothers, in 1935, discovered their eponymous phenomenological equation, $ \overrightarrow{J}= - \overrightarrow{A}/\xi^2$, so the London gauge $ \overrightarrow{\nabla}\cdot \overrightarrow{A}=0$ is actually current conservation $ \overrightarrow{\nabla}\cdot \overrightarrow{J}=0$.
