Balancing Redox Equations - Half Reactions
You are right for the half reaction for oxidation is:
$$\ce{Zn(s) \rightarrow Zn^{2+} + 2e^-}$$
Now, you have to write properly the  half reaction for reduction:
 $$\ce{2H^+(\mathrm{aq}) + 2e^- \rightarrow H_2(g)}$$

This means, that the electrons given by zinc are taken by ion hydrogen.
So, if we add the two half reactions:
$$\ce{Zn(s) + 2H^+(\mathrm{aq}) \rightarrow Zn^{2+}(\mathrm{aq}) +  H_2 (g)}$$
Notice, that ion chloride doesn't change its oxidation state. Its role is to keep the medium neutral electrically. So, we'll add two ions chloride to the two members of the equation:
$$\ce{Zn(s) + 2(H^+_{\mathrm{aq}} + Cl^{-}_{\mathrm{aq}}) \rightarrow (Zn^{2+}_{\mathrm{aq}} +2 Cl^{-}_{\mathrm{aq}}) + H_2(g)}$$
Remember, redox reaction are reactions of electrons exchange between an oxidant and an reducer.