Prove or disprove if $L_{1}$ is Turing-recognizable and $L_{2}$ is co-Turing-recognizable then $L_{1}\cap L_{2}$ is decidable I thought about these languages:
$$L_{1} = A_{TM} = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ accepts } w \big\}$$
$$L_{2} = \overline{HALT_{TM}} = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ doesn't halt on input } w \big\}$$

Their intersection:
$$L_{1}\cap L_{2}  = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ accepts and doesn't halt on input } w \big\}$$

If I assume that the intersection is decidable by TM $T$ so I can use $T$ to decide $A_{TM}$ and that's a contradiction.

Is it true?
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.