= Solution
The nonparametric <bootstrap sample> is a sample of size $n$ drawn independently with replacement from the <empirical distribution> $\widehat F_n=n^{-1}\sum_i\delta_{X_i}$, conditional on the observed data. To simulate it, take independent uniform $U_j$ and set $X_j^b=X_{1+\lfloor nU_j\rfloor}$, assigning $U_j=1$ to the last index if it occurs in a finite-precision generator.
Let $G_n^b$ be the conditional <cumulative distribution function> of $\sqrt n(\bar X_n^b-\bar X_n)$, and let $q_n(p)=\inf\{t:G_n^b(t)\geq p\}$. A basic <bootstrap confidence interval> is
$$
\boxed{C_n=[\bar X_n-q_n(1-\alpha/2)/\sqrt n,\ \bar X_n-q_n(\alpha/2)/\sqrt n].}
$$
In practice the conditional quantiles are estimated from many simulated bootstrap means; the asymptotic argument below describes exact bootstrap quantiles or a simulation count tending to infinity.
For the required uniform-convergence fact, given $\epsilon>0$ choose $a<b$ with $F(a)<\epsilon$ and $1-F(b)<\epsilon$. By continuity, choose a finite mesh on $[a,b]$ with successive increases of $F$ less than $\epsilon$. Pointwise convergence at the mesh points makes $|F_n-F|<\epsilon$ there for large $n$. Monotonicity sandwiches both functions between their neighbouring mesh values, so $|F_n(t)-F(t)|<2\epsilon$ throughout $[a,b]$. The same monotonicity and the chosen tail bounds control the two exterior rays by $2\epsilon$. Hence $\sup_t|F_n(t)-F(t)|\to0$.
The <bootstrap consistency theorem for the sample mean> states that, for iid observations with $0<\sigma^2<\infty$, the conditional law of $\sqrt n(\bar X_n^b-\bar X_n)$ converges weakly, in probability, to $N(0,\sigma^2)$. By continuity of that limiting distribution and the preceding monotonicity argument, the convergence of conditional distribution functions is uniform in probability. Since the limit is strictly increasing, $q_n(p)\to\sigma\Phi^{-1}(p)$ in probability for $0<p<1$. The coverage event is
$$
q_n(\alpha/2)\leq\sqrt n(\bar X_n-\mu)\leq q_n(1-\alpha/2).
$$
The <central limit theorem>, quantile convergence and <Slutsky theorem> therefore give $\boxed{\mathbb P^{\mathbb N}(\mu\in C_n)\to1-\alpha}$.
\b[Positive variance is needed for this exact coverage conclusion.] If the finite variance allowed in the wording is zero, all observations equal $\mu$ almost surely and the interval collapses to $\{\mu\}$, with coverage one. The bootstrap distribution is still consistent with the degenerate limit, but a nonrandomized usual interval cannot then have limiting coverage $1-\alpha$ for $0<\alpha<1$.
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