Suppose that algorithm $A$ makes at most $c\log n$ calls to the \textsf {choose} method. Create a deterministic version $A_i$ of $A$ for every possible outcome of these calls, and run $A_i$ to see if it accepts the given input. If one of the $A_i$'s accepts, then we accept the input. Otherwise, we reject it. If $A$ runs nondeterministically in $p(n)$ steps, then this deterministic version of $A$ will run in $p(n)2^{c\log n}=p(n)n^c$ steps, which is a polynomial.