= Solution
\b[Yes.] A <Cauchy sequence> in the supremum <norm> converges in the given complete space $B([0,1])$ to a bounded function $f$. Supremum-<norm> convergence is <uniform convergence>. The uniform limit of continuous functions is continuous: for a point $x_0$ and tolerance $\epsilon$, choose a continuous $f_n$ within $\epsilon/3$ of $f$ uniformly, then use its <continuity> at $x_0$ and the <triangle inequality>. Hence $f\in C([0,1])$. Thus $C([0,1])$ is a closed, complete subspace of $B([0,1])$.
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