Cuprous Chloride(WSDTY) does act as oxidizer, but there definitely is no intermediate Cu2O. HCl is not oxidizing at all in this place, although it plays an important role. What really happens is that the complexed Cu(+) ions are oxidized by O2 with the help of H(+) ions from the acid: 4Cu+(compl)+O2+4H+?>4Cu2+(compl)+2H2O. The H(+) ions are not oxidizing, but they help the oxidation to be carried out. Many oxidizers need H(+) for their oxidizing properties, e.g. permanganate, dichromate.

What I did was introduce the notation (compl). I did this to tell you that the reaction between copper (II) and copper metal indeed occurs, but not as simple ions. In reality, it is the complex ion CuCl2?4, which acts as oxidizer. The resulting ion is CuCl?2. This latter ion contains copper in the +1 oxidation state. When both ions are present, then even more complex things are formed, the deep brown mix-valency complexes, probably ClCu(μ-Cl)CuCl, which contains both copper in the +1 and copper in the +2 oxidation state (or two copper ions in the +1.5 oxidation state).

There is another thing I want to point out. You state that Cu2O is oxidized by HCl to CuCl2 and H2O. Just as an exercise, try to balance the following equation: Cu2O + HCl ---> CuCl2 + H2O

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