Finite State Machine that only accepts strings with equal number of 0's and 1's
The question is not really well-defined since there is no FSM that accepts your language $L$, so the existence of such an FSM implies anything – for example, $2+2=3$.

However, the intended solution was to notice that
$$ L \cap 0^*1^* = \{0^n1^n : n \geq 0\}. $$
Since the regular languages are closed under intersection, this shows that an FSM for $L$ can be converted to one of $\{0^n1^n : n \geq 0\}$.

Your attempt claims that an FSM accepting $L$ must have a counter. This is not a formal statement: it has no formal content. It's just intuition. A PDA for $L$ also doesn't have a counter; it has a stack. Counter is not a well-defined technical notion.