Since the voltage drop ratio is relatively low and the output current requirements also low, use a linear regulator. 3.3 V is a common voltage, so there are many fixed linear regulators available at that voltage.
These things have only three pins and are very simple to use. The pins are the input voltage, ground, and the output voltage. You will also need a 1 µF or so ceramic cap between input and ground, and between output and ground.
You are dropping (4.5 V) - (3.3 V) = 1.2 V. You have to be careful to choose a regulator that can work with that headroom. These are often called LDOs (Low DropOut).
The efficiency from the voltage drop will be 73%, plus a little more loss for the quiescient current. At only 400 µA output, the overall wasted power will be very small.
Also take a look at the quiescient current spec. For some linear regulators, that would add significantly to your 400 µA figure. Others work with only a few µA.
Take a look at the MCP1700 series, but there are many many others that would be fine too.
Some older LDOs are not "0 ESR output cap stable". Simply stay away from them. They were designed before the era of small and cheap ceramic capacitors that could do a few µF.
The MCP1700 series I mentioned is 0 ESR output stable, requires a maximum of 350 mV headroom, has only 4 µA quiescient current, and can deliver up to 250 mA. These are my "jellybean" LDOs, meaning that's what I use unless there is a good reason not to. I don't see one in this case.