Using a boost converter as a laser driver for high power diode
Well, sorry about that ;) pretty much everything in the schematic is wrong. Where is the inductor? And the input/output decoupling? C3/C4 will block output voltage. There should be an example schematic in the datasheet.
A synchronous converter (without diode) would offer better efficiency which should be important when using battery power.
Note that the input power of a boost converter is equal to output power plus losses, therefore input current will be:
\$ Iin = \frac{1}{Efficiency} Iout \frac{Vout}{Vin}  \$
With 90% efficiency, 4A 5V out, 3V in, this gives an input current of 7A. This is way above what a standard 18650 LiIon battery can provide so you will need a high current capable battery, as used in e-cigarettes for example.
If you used two lithium cells instead of one, you could use a buck converter, whose input current follows the same equation... but in this case Vin&gt;Vout which means Iin will be much lower.
With 90% efficiency, 4A 5V out, 6V in, this gives an input current of 3.7A, much easier on the batteries. It also has higher efficiency due to less RI^2 losses. So, only use a boost if you absolutely want to use one cell to make it small.
You could use a boost LED driver chip, or a buck LED driver chip if you use 2 cells, which will regulate current. This would be mots convenient. You can use a voltage output DC-DC converter too but then it's wasteful to put a resistor in series with the laser diode. You can make a voltage output buck converter regulate current by using a current sense amp and a shunt resistor in series with the laser diode, and an opamp to amplify the voltage on it to make the desired current correspond to the feedback voltage on the DC-DC converter, but this requires some attention to frequency compensation.