I eventually used a trial and error method to solve it:
This let me set B1 = 4 and hence solve the puzzle. But is there a generic method I could have applied here to solve this?
Just look at the remaining 4s. If B4 is a 4, then so is C2 and then so is F1. But F1 and B4 block both 4 options in the right-center section, so the left-center 4 must be C3. The rest of the solution thereafter is trivial.
You can prove:
D4=4, otherwise D4,E4,D5,E5 isn't unique.
This is another trial and error attempt I did:
If you choose F1=4, this will reduce the options of the '4' in rows 2 and 3 to be at column 'C'. C2=C3=4. Then, the '4' of the first block can't be at C2 and must be at B1.
I would generally start looking at forced chains of cells with 2 options left.
In this particular case my first idea would be to follow the chain
A3 -> A6 -> B6 -> B1 -> C2 -> C3
This means:
either A3 is a 2 or if it's a 6 (following the forced chain) C3 is a 2.
Thus B4 and F3 cannot be 2.
And next F needs a 2 in top block so D2 can't be a 2
Similar chaing reasoning can be used to see if a cell with 3 options is part of such a chain as well to reduce it to only 2 options instead:
In this case (after previous steps) I would look at F2 starting the chain from F3:
F3 = 4 -> F2 = 4
F3 = 6 -> A3 = 2 -> C3 = 4 -> C2 = 2 -> F2 = 6
=> F2 cannot be 2
Which then leads us to F1 = 2 and solving the rest trivially.
I do want to point out though that the above steps took longer than your trial and error solution.
Which one is the best way to solve it depends largely on what you want.
I personally don't like guessing without reasoning so will only fall back on trial and error if all other options are exhausted ... or at least untill it takes too long to spot any of those other options. Nothing wrong with at least trying something instead of staring at a really hard puzzle without progress.
Final note: sometimes when just trying something randomly makes it easier to spot other forced hints even if your initial guess didn't directly lead to reducing any options.
F1 must be 2
Why?
Let's say we put in a 4 at the orange marked cell F1. Then B1 must be a 2; C2 a 4; C3 a 2; A3 a 6; and B4 a 4.
However, we now have a 4 at F1 and a 4 at B4, eliminating the 4 from the entire middle-right block (F3/D4), which isn't correct. Because of that, we can conclude that F1 cannot be a 4 and must be a 2. After that it should be trivial to continue.
