Boiling of water does not involve temperature change, so why is there a change in internal energy during the process?
The change in internal energy $U$ is dependent on the heat transferred $q$ and the work done $w$. 

$$\Delta U = q + w$$

$\Delta U = 0$ in two cases: when $q=-w$ and when $q=w=0$.

The heat transferred is not necessarily zero if there is no temperature change. The most common expression of heat transfer is $q=C\Delta T$, where $C$ is the heat capacity of the system and $\Delta T$ is a temperature change. We can also write $q=ms\Delta T$, where $m$ is the mass of the system and $s$ is its specific heat capacity (per unit mass). However, heat is being transferred during boiling (else how does the liquid boil?). We need a new equation for heat that does not require temperature.

In a constant pressure system (most boiling happens in a constant pressure,  open to the atmosphere, system), change in enthalpy is equal to $q$ since all of the work and other pressure-volume terms cancel. The enthalpy change of vaporization is usually given in $\mathrm{kJ}$ per mole or unit mass. Now we have a new equation for heat:

$$q_{vap}=m\Delta_{vap}H$$

Let's say we are boiling 100.0 g of water. Water's enthalpy of vaoprization is $2257\ \mathrm{kJ/kg}$. Thus the heat transfer is:

$$q=(0.1000\ \mathrm{kg})(2257\ \mathrm{kJ/kg})=225.7\ \mathrm{kJ}$$
Note that in this case, the work done is also not zero. There is a volume change of the system. Let's take our 100 grams of water. 

$$V=0.1000\ \mathrm{kg}\times \dfrac{\mathrm{m^3}}{958.4\ \mathrm{kg}}=1.040\times 10^{-4}\ \mathrm{m^3}$$

At the boiling point, the density is $958.4\ \mathrm{kg/m^3}=0.9584\ \mathrm{kg/L}$. Using the ideal gas law (water vapor is reasonably ideal at atmospheric pressure and its boiling point), we can determine the volume of 100.0 g of water vapor, given that 

$$100.0\ \mathrm{g}\ \ce{H2O}\times \dfrac{1\ \mathrm{mol}\ \ce{H2O}}{18.02\ \mathrm{g}\ \ce{H2O}}=5.549\ \mathrm{mol}\ \ce{H2O}$$

$$V=\dfrac{nRT}{P}=\dfrac{(5.549\ \mathrm{mol})(0.080257\ \mathrm{\frac{L\cdot atm}{K\cdot mol}})(373.15\ \mathrm{K})}{1\ \mathrm{atm}}=169.9\ \mathrm{L}$$

We can now calculate work. So that we get joules out, I am going to convert pressure to pascals and volume to $\mathrm{m^3}$.
$$P=1\ \mathrm{atm}=101,325\ \mathrm{Pa}\\
V_{gas} = 169.9\mathrm{\ L} =0.1699\ \mathrm{m^3}\\
\Delta V= 0.1699\ \mathrm{m^3} - 1.040\times 10^{-4}\ \mathrm{m^3} =0.1698\ \mathrm{m^3}\\
w=-P\Delta V=-101325\ \mathrm{Pa}\times 0.1698\ \mathrm{m^3}=-17.20\ \mathrm{kJ}$$

Now we calculated $\Delta U$:

$$\Delta U = q +w = 225.7\ \mathrm{kJ} - 17.20\ \mathrm{kJ} = 208.5\ \mathrm{kJ}$$