Prove or disprove if $L_{1}$ is Turing-recognizable and $L_{2}$ is co-Turing-recognizable then $L_{1}\cap L_{2}$ is decidable
Your proof is wrong. Note if $M$ accepts $w$, it must halt on $w$, so $L_1\cap L_2$ is the empty set, which is of course decidable. 

To disprove the statement, you can set $L_2$ to be $\Sigma^*$, i.e. the language containing all strings. Since $\emptyset$ is Turing recognizable, $L_2$ is co-Turing recognizable. However, $L_1\cap L_2=L_1$ is undecidable.