Number of states in minimal Non Deterministic Finite Automata
But will not this NFA accept every string over the alphabet set?


It will not accept any strings except the empty one.
Since there are no transitions, the transition function $\delta(q_0,a) = \emptyset$ for all symbols $a$ of the input alphabet $\Sigma$, thus the extended transition function $\hat{\delta}(q_0,s) = \emptyset$ for all non-empty strings over $\Sigma$.

The set of accepting states $\{q_0\}$ and $\hat{\delta}(q_0,s)$ for all non-empty strings $s$ are disjoint, hence the NFA doesn't accept any non-empty strings.

As you already know this NFA accepts the empty string, and we just showed that it accepts only the empty string.

Remark: if it was a deterministic automaton you'd need 2 states to do the same thing. You could "cheat" and not show the 'dead state', but still it would be a 2 state DFA. With NFAs rules are a little bit different, since instead of states you use sets of states and having your NFA in no state at all (the empty set of states) is perfectly normal abstraction.