Does the diameter of the earth decrease because of soil subsiding due to irrigation and rainfall over the years?
Disclaimer: this is probably TLWR, but you can absolutely apply some physical reasoning to arrive at the answer.

Changes in surface height of the earth are governed by the conservation of mass, i.e. that if mass ($M$) is removed in one location (everglades soil erosion), then an equal amount of mass must be gained at another location (perhaps accretion on the seafloor).The total change in the earths mass is zero $$\Delta M = M_1-M_2=0$$
$$-M_1=M_2$$. 

Mass may be broken up into the volume ($V=\Delta x \Delta y \Delta z$) multiplied by density ($\rho$), so $M=\rho V$. In terms of the conservation of mass equation from above, then:
$$-\rho_{1} \Delta x_1 \Delta y_1 \Delta z_1 = \rho_{2} \Delta x_2 \Delta y_2 \Delta z_{2}$$

The place where erosion occurs is described by the left side (with subscript 1). The place where accretion occurs is described by the right side (with subscript 2). This equation says the total mass lost by area 1, must equal the total mass gained by area 2. Now consider the following two situations.

1) The transported sediment does not undergo any change in density. If the density does not change, you have $\rho_1=\rho_2$. Let's also say the eroding area ($\Delta x_1 \Delta y_1$) equals the accreting area ($\Delta x_2 \Delta y_2$) for simplicity. Then all equal terms in the conservation of mass cancel out and you're left with
$$-\Delta z_{1} = \Delta z_{2}$$

2) The transported sediment does undergo a change in density. There's a host of reasons why density can change. If the density changes then we can't let the density terms cancel in the conservation of mass. Lets still say the eroding area and accreting area are equal. The conservation of mass becomes 
$$-\rho_1 \Delta z_1 = \rho_2 \Delta z_2$$
or conversely 
$$-\Delta z_1=\frac{\rho_2}{\rho_1}\Delta z_2$$
This says that the amount of change in height at location 1 is not equal to the amount of change at location 2 (sure they're proportional, but not definitely not exactly equal).

Putting it all together to answer the question, the average earth radius (r)  is the average of all heights (z) with respect to the center of the earth $$r=\frac{z_1+z_2+...+z_n}{n}$$

Try this all out with arbitrary numbers to convince yourself. Add the changes $\Delta z_1$ to $z_1$, and $\Delta z_2$ to $z_2$
$$r_{before}=\frac{z_1 + z_2 + z_3}{3}$$ 
$$r_{after}=\frac{(z_1+\Delta z_1)+(z_2+\Delta z_2)+z_3}{3}$$
Rearrange $r_{after}$ to be
$$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1+\Delta z_2)}{3}$$

Case 1 If density doesn't change (i.e. $-\Delta z_1=\Delta z_2$), then $$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1-\Delta z_1)}{3}$$
The changes cancel out and $r_{before}=r_{after}$. The earths radius does not at all change.

Case 2 If the density changes, $r_{after}$ becomes
$$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1-\frac{\rho_2}{\rho_1}\Delta z_1)}{3}$$
and you can see that $r_{after}$ does not equal $r_{before}$.