That follows from Rice's theorem:
Rice's Theorem.
Let S be a set of languages that is nontrivial, meaning
- there exists a Turing machine that recognizes a language in S
- there exists a Turing machine that recognizes a language not in S
Then, it is undecidable to determine whether the language decided by an arbitrary Turing machine lies in S.
[Copied from Wikipedia]
The language $\{00,01\}$ is nontrivial, thus $L$ is not decidable.
Hint: It is also easy to prove that directly, not using Rice's theorem. Let $M$ be a Turing machine. Consider Turing machine $M'$ defined as
if $x\in\{00,\,01\}$ then
run $M$ on empty tape
accept (if $M$ halts)
otherwise, reject