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I know that $\Delta S \geq \int_a^b \frac{\delta Q}{T}$ holds in general. When the system (with its heat sources which interacts only with it) is isolated, I read that, since $\delta Q=0$, then for the equation above $\Delta S \geq 0$. But the $\delta Q$ in the integral are the amounts of heat that are exchanged between, for example, a gas and each of the heat sources, and T is the temperature of each source which of course is different for each one of them. Then why in an isolated system each $\delta Q$ must be zero? Isn't zero the $Q$ exchanged between the system and the outside?

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Then why in an isolated system each must be zero? Isn't zero the exchanged between the system and the outside?

By definition, an isolated system is one in which there is no mass transfer and no transfer of energy in the form of work or heat between the system and its surroundings. So, by definition, for an isolated system $\delta Q=0$

The fact that $\delta Q=$ does not mean there can be no entropy change in an isolated system. Although the entropy change is defined for a reversible transfer of heat you can devise any convenient reversible path between two equilibrium states and calculate the change in entropy of the system using the equation

$$\Delta S_{sys}=\int\frac{\delta Q_{rev}}{T}$$

because entropy is a state function independent of the path.

The classic example is that of an isolated system in which an ideal gas expands irreversibly into a vacuum. See the explanation in my answer here:Entropy change in the free expansion of a gas

Hope this helps.

Bob D
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