How can I prove the minimum pumping length of finite language?
You explanation sounds good (you want to change "smaller than" in "smaller than or equal to").

Suppose towards a contradiction that the pumping length was $k \le 3$ then, since 
$w=000$ is a word of length at least $k$ in $L$, you could invoke the pumping lemma to conclude that $w$ can be written as $w=xyz$, with $|y|&gt;0$, in such a way that $xy^ik \in L$ for any choice of $i \ge 0$.

This is a contradiction since the choice $i=4$ yields a word $w'$ of length $|w'| = |x| + 4 |y| + |z| \ge 4$ for which you simultaneously have $w' \in L$ (by the Pumping Lemma) and $w' \not\in L$ (by the definition of $L$).

In addition, to show that $k'=4$ is the smallest possible pumping length, you need to argue that $4$ is a valid choice for the pumping length. 
In your case this is trivial as the pumping lemma gives you a property over the set of all words in $L$ of length at least $k'$. For $k'=4$ this set is empty, which makes the statement trivially true.