L={⟨M⟩: M is a DFA and for each string in L(M) the number of 1s is more than or equal to the number of 0s }
T = "On input where M is encoded DFA"
1. Construct another DFA D such that L(D)={x|x has more or equal 1s than 0s}
2. For each input x
a. if( x is accepted by DFA M){
if(accepted by D){
"accepted" and return;
}
3. "rejected"
Is this proof true?
No, this proof is not correct. You can't iterate through all inputs $x\in \Sigma^*$ since it would take you "infinite time".
The correct way to do this is to construct the complement of $D$ (as a pushdown automaton! as @Steven mentioned), which we will call $D^c$, then construct the intersection PDA $D^c\cap M$ (notice that this can be done since $M$ is a DFA), and test if its language is empty. If it is, you can be sure that $L(M)\subseteq L(D)$, hence all words in $M$ have more $1$'s than $0$'s
