This is a grossly oversimplified explanation, since it assumes that there is a sudden discontinuity between the portion of a conductor that suffers skin effect and the portion that does not:
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Suppose we have a 1cm diameter (0.5cm radius) wire. Due to skin effect, conduction only takes place in the part of the wire from 0.4 cm out to the periphery. The effective cross sectional area that conducts current is pi(.5^2) - pi(.4^2). = .25pi - .16pi = .09pi. Now suppose we use a wire whose radius is equal to the skin depth (.1cm). The entire cross sectional area of this smaller wire conducts current. The cross sectional area of this wire is pi(.1^2pi) = .01pi. It takes 9 of the smaller wires to produce the same cross sectional area as the effective cross sectional area of the larger conductor. We can fit much more than 9 of the smalller wires inside the actual cross sectional area of the larger conductor. Therefore the effective cross sectional area of the stranded wire consisting of as many small wires as will fit inside the cross sectional area of the larger conductor will have a much lower effective high frequency resistance than the single large diameter wire.
Regards,
Kral