You did everything right, and you're almost there. Your last equation becomes

$$\sum_lc_l\sum_md_m\delta(n-l-m)\tag{1}$$

Due to the Kronecker delta the sum over $m$ reduces to a single element:

$$\sum_md_m\delta(n-l-m)=d_{n-l}\tag{2}$$

because $\delta(n-l-m)$ is non-zero for $m=n-l$ and zero otherwise. Plugging $(2)$ into $(1)$ gives the desired result:

$$\sum_lc_ld_{n-l}$$

You did everything right, and you're almost there. Your last equation becomes

$$\sum_lc_l\sum_md_m\delta[n-l-m]\tag{1}$$

Due to the Kronecker delta the sum over $m$ reduces to a single element:

$$\sum_md_m\delta[n-l-m]=d_{n-l}\tag{2}$$

because $\delta[n-l-m]$ is non-zero for $m=n-l$ and zero otherwise. Plugging $(2)$ into $(1)$ gives the desired result:

$$\sum_lc_ld_{n-l}$$