= Solution
Let $\Delta_n=\sup_{\theta\in\Theta}|Q_n(\theta)-Q(\theta)|$. For any $\varepsilon>0$, the set
$$
F_\varepsilon=\{\theta\in\Theta:\|\theta-\theta_0\|\geq\varepsilon\}
$$
is <compact>. If it is empty there is nothing to prove. Otherwise <continuity> and the unique minimum give a strictly positive separation gap
$$
d_\varepsilon=\min_{\theta\in F_\varepsilon}\{Q(\theta)-Q(\theta_0)\}>0.
$$
Take any measurable attained <minimizer> of $Q_n$. Its defining inequality gives
$$
Q(\widehat\theta_n)\leq Q_n(\widehat\theta_n)+\Delta_n\leq Q_n(\theta_0)+\Delta_n\leq Q(\theta_0)+2\Delta_n.
$$
Consequently
$$
\boxed{\Pr(\|\widehat\theta_n-\theta_0\|\geq\varepsilon)\leq\Pr(\Delta_n\geq d_\varepsilon/2)\longrightarrow0.}
$$
This proves <argmin consistency under uniform convergence in probability>. No <continuity> of $Q_n$ is needed once the <minimizer> exists; compactness and <continuity> concern the deterministic separation gap. The <uniform convergence in probability> assumption controls all candidate parameters, including the random <minimizer>.
For the <estimating equation>, fix $\varepsilon>0$ and put $a=\theta_0-\varepsilon$, $b=\theta_0+\varepsilon$. The prescribed signs make $\eta=\tfrac12\min\{-S(a),S(b)\}>0$. Pointwise <convergence in probability> at just these two points implies
$$
\Pr(S_n(a)\geq0)\leq\Pr(|S_n(a)-S(a)|\geq\eta)\longrightarrow0,
$$
and similarly $\Pr(S_n(b)\leq0)\to0$. With <probability> tending to one, $S_n(a)<0<S_n(b)$. The <intermediate value theorem> then gives a zero inside $(a,b)$, and uniqueness identifies it with $\widehat\theta_n$. Hence
$$
\boxed{\Pr(|\widehat\theta_n-\theta_0|\geq\varepsilon)\leq\Pr(S_n(a)\geq0)+\Pr(S_n(b)\leq0)\longrightarrow0.}
$$
This is <consistency of a uniquely bracketed zero>. It needs no monotonicity, no <continuity> of the limit $S$, and no uniform convergence of the $S_n$. The bracket interval must lie in the domain of $S_n$: read literally, the printed sign condition for every positive $\varepsilon$ puts every real point into $\Theta$, so this requirement is satisfied. More generally it is enough to have an interval about $\theta_0$ and sign brackets arbitrarily close to it.
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