LT3652 Inductor Selection
I believe there is an error in your calculations and the actual value is 2x your calculated value.

The datasheet gives

$$
L = \frac{10\cdot R_{sense}}{\frac{\Delta I_L}{I_{CHG(MAX)}}} \cdot V_{BAT(FLT)} \cdot \left(1-\frac{V_{BAT(FLT)}}{V_{IN(MAX)}}\right) \text{(uH)}
$$

where \$\Delta I_L = 0.25 \cdot I_{CHG(MAX)} = 0.5 \text{A}\$.

With \$R_{SENSE} = 0.05\text{ } \Omega\$, \$V_{BAT(FLT)} = 4.2 \text{V}\$, \$I_{CHG(MAX)} = 2.0\text{A}\$, \$V_{IN(MAX)} = 9.94\text{V}\$, we have:

$$
L = \frac{10\cdot 0.05}{\frac{0.5}{2}} \cdot 4.2 \cdot \left(1-\frac{4.2}{9.94}\right) = 4.85 \text{ uH}
$$

if my calculator is correct. So a 5.1 uH inductor would probably be a good choice. On page 22 of the datasheet there is a 1 cell LiFePO4 reference design, which has a 5.6 uH inductor (similar to your design). However, as your maximum input voltage is lower, your inductor can be smaller. Of course choosing a larger inductor (within reason) will probably not hurt the circuit, and improve the ripple. So consider 5.1 uH as more of a lower bound, anything between 5 uH and 10 uH is probably appropriate.

Of course, you should build a prototype with your chosen inductor value, and then measure voltage ripple, battery charging current, and so on, and make sure that these look tolerable for your application.