Shouldn't change in entropy always be 0? We know that entropy is defined as $$dS=\frac{dQ}T$$ Now in any isolated system, by definition $dQ=0$.  So should entropy change in any isolated process or system always be 0?
 A: You have misunderstood the meaning of the symbol $dq$ in your statement about entropy. The relationship between small entropy change $dS$ and heat entering the system $dq$ is
$$
dS \ge \frac{dq}{T}.
$$
In order to get an equality in this relationship the heat transfer process must be reversible (in the thermodynamic sense). This can be expressed
$$
dS = \frac{dq_{\rm rev}}{T}
$$
and any finite entropy change is given by
$$
\Delta S = \int \frac{ dq_{\rm rev} }{T}.
$$
The way to read the right hand side of this equation is "the integral of the heat that would be transferred if the change of state were to be carried out via a reversible path, divided by temperature."
In the case of an isolated system one has $dq = 0$. From this we can deduce that
$$
dS \ge 0
$$
and therefore
$$
\Delta S \ge 0
$$
for any process. The case of an isolated system undergoing an irreversible process (e.g. free expansion) results in $\Delta S > 0$. We can calculate this $\Delta S$ using
$$
\Delta S = \int_i^f \frac{ dq_{\rm rev} }{T}
$$
because, as I said before, the $dq_{\rm rev}$ here is not the heat transfer in the process the system actually followed (e.g. free expansion); it is the heat that would be transferred if the system were to move between the same two states $i$ and $f$ by a reversible process.
A: Your point of misunderstanding is what constitutes an isolated system. Also, your equation is wrong.
$$dS=  \frac{dQ}T $$
with no integral. If you want to integrate to get total entropy, you have to integrate both sides.
So yes, we could isolate some gas at complete equilibrium and hold it at constant pressure, temperature, and volume, and we would find that the change in entropy of that system was zero.
However, if we drop an ice cube into that gas and isolate our new ice-gas system, we have
$$dS_{gas} + dS_{ice} = \frac{dQ_{gas}}{T_{gas}} + \frac{dQ_{ice}}{T_{ice}} \geq 0 $$
The ice and the gas do work on one another as they equilibrate, and the entropy of each subsystem will increase or decrease accordingly, while the total entropy of our ice-gas system stays the same or increases.
