Understanding the guarding circuit in an instrumentation amplifier
I have an explanation which tells how U1C affects the CMRR. I have also made PDF file of the following explanation and uploaded in my drive which can be found here. The PDF file is easy to read than the following

Explanation:

The net circuit can be divided into two parts. The first part consists of U1B, U1C and U1D. The second part consists of U1A (the usual differential amplifier circuit).

Consider the first part, the part which takes the input and increases the CMRR. At the op-amp U1C, the voltage at +Ve terminal and –Ve terminal are equal. So both are zero volatage. So it means the voltage V is zero in the following circuit (first half of the circuit):



As input bias current is almost zero, op-amp draws almost 0 current. So by KCL law, from the above circuit, we get:

$$\frac{0-V_2}{R_1} =  \frac{V_1-0}{R_2}$$   

As \$R_1\$ becomes equal to \$R_2\$, we get \$V_2 = -V_1\$. Remember that these \$V_1\$ and \$V_2\$ are inputs to differential amplifier. So the difference of the input signal increases and their average becomes zero as \$R_1\$ becomes equal to \$R_2\$. And this happens with help of the op-amp U1C (remember this op-amp helps by drawing zero current and it making V=0).

More explanation:

Let us analyze the upper left op-amp U1D. Let the output of op-amp U1C be \$x\$ volts. So the op-amp U1C has been modeled as separate voltage source as shown below: 



The output of the above op-amp U1D is:

$$V_2 = {\frac{1+R_7}{R_9}}\cdot(V_b-x)$$

This actually our \$V_2\$. Similarly, the output of U1B is:

$$V_1 = {\frac{1+R_8}{R_{10}}}\cdot(V_a-x)$$

This actually our \$V_1\$.

As \${V_1}\cdot{R_1}={-V_2}\cdot{R_2}\$ and \$R_1=R_2\$  (from figure), we get:

$$x = V_a+V_b$$

So

$$V_1= {\frac{1+R_8}{R_{10}}}\cdot(\frac{V_a}{2}-\frac{V_b}{2})$$

and

$$V_2= {\frac{1+R_7}{R_9}}\cdot(\frac{V_b}{2}-\frac{V_a}{2})$$

As \$R_8=R_7\$ and \$R_9=R_{10}\$, we get:

$$V_1=V_a-V_b$$

and

$$V_2=V_b-V_a$$  

Now consider the following circuit



From the above circuit

$$CMRR=\frac{V_1-V_2}{\frac{V_1+V_2}{2}}$$

From the above values we get:

$$CMRR = \frac{2(V_a-V_b)}{0}=\infty$$

CMMR equal infinity is just theory as resistance cant be accurately equal. They have tolerance level.   

Note: CMRR is the value for the differential amplifier considering the effects of U1B, U1C and U1D.