Copper is used to avoid lithium-halogen exchange. Additional information on that reaction can be found in this discussion. Organolithiums will often, instead of acting as a nucleophile, form a new organolithium: $$\ce{RLi + R'X <=> RX + R'Li}$$ Using $\ce{Cu(I)}$ allows the formation of an organocuprate, which acts as a nucleophile, allowing alkylation rather than metallation: $$\ce{2RLi + CuX -> LiR2Cu + LiX}$$ $$\ce{LiR2Cu + R'X -> R'R + RCu + LiX}$$ I'm not sure if silver could be used instead; I'm not seeing any refrencereference for it, but I may just not be searching the right terms. The this 2005 paper pageWikipedia page on organosilver chemistry isn't terribly extensive, but I'm guessing stoichiometric use of silver, while not terribly expensive as transition metals go, would certainly be more so than analogous use of copper.
Update:
It seems that analogous chemistry cannot be performed with silver or gold. Lithium diorganoargentates ($\ce{LiR2Ag}$) are apparently too unstable. Lithium diorganoaurate ($\ce{LiR2Au}$) complexes are apparently either not good nucleophiles, or the triorganogold(III) ($\ce{R2AuR'}$) intermediate complexes are too stable to perform similar chemistry. This is per this 2005 paper. A fair amount of gold chemistry has come out since then, so I'm not sure how much of this has changed since.
Second update (though I should have answered this to begin with):
Also, I'd like to add that this is not an $\text{S}_\text{N}2$ reaction. The organocuprate undergoes oxidative addition into the $\ce{R'X}$ bond. After that, it loses the $\ce{X^-}$ ligand to form $\ce{LiX}$ and $\ce{R2CuR'}$, then undergoes reductive elimination to form $\ce{R'R + CuR}$.