Prove no DFA with four states can accept L={111}
This depends on your favorite definition for DFAs. If your definition allows the transition function to be undefined on some state-symbol pairs $(q, a)$ then your DFA in your comment does the trick. Otherwise, you would need a fifth state (a sink) that is not accepting and catches all missing transitions.
To show that there is no DFA with less than five states accepting $L = \{111\}$, we can check the Myhill-Nerode equivalence classes of $L$ which are

$[\varepsilon]_L = \{\varepsilon\}$,
$[1]_L = \{1\}$,
$[11]_L = \{11\}$,
$[111]_L = \{111\}$,
$[0]_L = \{w \in \{0, 1\}^\ast \mid w \notin \{\varepsilon, 1, 11, 111\}\}$.

(To show that these are indeed the five classes is left as exercise.)
The equivalence classes have a 1:1 correspondence to the unique minimal DFA of the latter definition (with transitions for all state-symbol pairs).