I tried to prove by contradiction.
$L_{1}$ is undecidable and $L_{2}$ is finite language then $\overline{L_{1}}\cap \overline{L_{2}}$ is decidable. $$L_{1} = \overline{HALT_{TM}} = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ doesn't halt on input } w \big\}$$ $$L_{2}=\emptyset$$ then, $$\overline L_{1}={HALT_{TM}}=\big\{ \langle M, w \rangle \mid M\text{ is TM and }M\text{ halts on input }w \big\}$$ $$\overline L_{2}=\Sigma^{*}$$ The intersection, $\overline{L_{1}}\cap \overline{L_{2}}$ is $\overline L_{1}={HALT_{TM}}$ is decidable and this is the contradiction.
Is it true?
Thanks.