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?