For completeness, decided to add the optimality proof.
Assume the opposite - i.e. you can cover the board with tetraminos, such that there is not a pond of size larger than 3.
Now consider the 4 squares a7, a8, b7, b8. At least one of them should be covered by a tetramino.
Case 1. Only one square is covered. Now there are basically 3 different subcases to consider, first two of which are straightforward:
For the third case, when we have a tetramino on b8, c8, c7, d7, has 2 subcases. The first one leads us to (up to symmetry)
and the second one leads us to
Case 2. 2 squares are covered. WLOS let them be b7 and b8. Now using the observation above it is easy to see that we should have tetraminos on the following 2 places:
However, now the square d6 is an issue and once again we get a contradiction.
Case 3. 3 squares are covered. If a8 is covered, then (up to symmetry) we get:
Otherwise WLOG we have a tetramino on a7, b7, b8, c8. Also because of the observation above the square c6 should be covered as well. Similarly, a2, b1, b2, c3 should be covered by tetraminos. Now it is easy to see that a4, a5, a6, b6 can not be covered by tetraminos, so once again we get an island of size 4. Contradiction.