Proof.
Consider any prime \mathfrak q' \subset S which corresponds to a point of \mathop{\mathrm{Spec}}(S_{\mathfrak p}). This means that \mathfrak p' = R \cap \mathfrak q' is contained in \mathfrak p. Here is a picture
\xymatrix{ \mathfrak q' \ar@{-}[d] \ar@{-}[r] & ? \ar@{-}[r] \ar@{-}[d] & S \ar@{-}[d] \\ \mathfrak p' \ar@{-}[r] & \mathfrak p \ar@{-}[r] & R }
Assume (1) and (2)(a). By going up there exists a prime \mathfrak q'' \subset S with \mathfrak q' \subset \mathfrak q'' and \mathfrak q'' lying over \mathfrak p. By the uniqueness of \mathfrak q we conclude that \mathfrak q'' = \mathfrak q. In other words \mathfrak q' defines a point of \mathop{\mathrm{Spec}}(S_{\mathfrak q}).
Assume (1) and (2)(b). By going down there exists a prime \mathfrak q'' \subset \mathfrak q lying over \mathfrak p'. By the uniqueness of primes lying over \mathfrak p' we see that \mathfrak q' = \mathfrak q''. In other words \mathfrak q' defines a point of \mathop{\mathrm{Spec}}(S_{\mathfrak q}).
In both cases we conclude that the map \mathop{\mathrm{Spec}}(S_{\mathfrak q}) \to \mathop{\mathrm{Spec}}(S_{\mathfrak p}) is bijective. Clearly this means all the elements of S - \mathfrak q are all invertible in S_{\mathfrak p}, in other words S_{\mathfrak p} = S_{\mathfrak q}.
\square
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