$O$ ctomino

Let's look at all possible centers of $O$ octominos.

They are all within the highlighted inner $8\times8$ square, since an $O$ octomino's center (for obvious reasons) can't be on any of the pink colored unit squares.

Our goal is to add a minimum number of stars to specific unit squares in a way that in the end we can not put an $O$ octomino inside the $10\times10$ square without it containing at least one star. In other words with each added star we eliminate possible $O$ octomino centers, until there aren't any remaining.

We can see that adding a star removes the possible centers around the star (since any $O$ octomino with a center around the star would cover the star):

The eliminated squares form an $O$ octomino, (meaning adding a star is equivalent to eliminating an $O$ octomino from the possible centers)

With all the information so far we can reframe our question in the following way:

How many $O$ octominos are needed to fully cover an $8\times8$ board?

For each square with C in it (on the picture above), there is only one $O$ octomino that covers it, so we can be certain that these octominos must be used. This means $4$ octominos so far.

Now to finish off the problem look at the yellow squares. It's clear that no $O$ octomino contains more than $2$ of them.

Since there are $16$ yellow squares, we know that we need at least $16/2 = 8$ more $O$ octominos to fill the board, which means we need at least 12 in total, this also means we need at least $12$ stars.

(In the middlle, you can actually place an octomino that covers 3 yellow squares, but that would mean we have 6*2+3 = 15 yellow squares covered, after placing 7 octominos, so again we would need at least 8).

We can show that $12$ stars are enough. Here's an example: