The gas temperature is the same as the temperature of the reservoir
This is exactly why there is energy exchange.
You should think in small steps: The volume is expanding, but the change occurs slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange.
See Isothermal process for more details.
As stated in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.
On the other hand, there's a simple experiment which shows that for nearly ideal gases, such as air at standard conditions, $U$ is a function of only temperature $T$. (It is therefore assumed for ideal gases, that $U$ is exactly a function of $T$.)
In the experiment, we have an adiabatic container with a wall that divides it in two compartments: one side is filled with air at atmospheric pressure, and the other is a vacuum.
Then, we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, we observe the temperature practically stays constant.
Now let's look at the whole process. Since the gas is thermally insulated, it doesn't exchange heat with the surroundings: $$Q=0.$$
The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$
So, by the conservation of energy: $$\Delta U = Q+W=0.$$
The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So, we conclude internal energy is independent of the volume for ideal gases.
Best Answer
The ideal gas law says that for a fixed number of molecules in a gas $$\frac{PV}{T}=\text{constant}.$$ If the process is isothermal, as you specify, then $T$ is constant and we can write $${PV}=\text{constant}.$$ This means that in an isothermal process, both pressure and volume must change in order for any change in state to occur.
Now, you mention the equation $$dU=dQ=nC_V∆T.$$ The subscript on the specific heat value, $C_V$, means that this applies for constant volume. So you have a situation in which this calculation does not apply. Also, notice that if $V$ is constant, then $$\frac{P}{T}=\text{constant}$$ and no work is done, because work requires a change in volume. This is incompatible with an isothermal process.
What you should realize is that this is not the only method of calculating heat addition or removal. Also, the first law says $$\text{d}U=Q_{in}+\text{d}W$$ In this particular form of the statement, $Q_{in}$ positive if heat flows into a system and negative if heat is removed from the system, and $\text{d}W$ is positive if work is done on the system and negative if work is done by the system.
In an isothermal process, with both $P$ and $V$ changing, but $T$ being constant, work is being done on (or by) the system so the internal energy must change unless heat is removed (or added to) the system, respectively.
In thermodynamic processes you must be careful to understand when you are applying a specialized condition (heat in a constant volume situation, which is incompatible with isothermal processes) rather than a general condition (first law of thermodynamics).