= Solution
The expected <check loss> is finite because $\rho_\tau(u)\leq |u|$. It is also Lipschitz in its location argument, with constant at most one. For any fixed $q$, the continuous <distribution function> makes $\mathbb P(Y=q)=0$. The derivative of $\rho_\tau(Y-q)$ with respect to $q$, away from that null event, is $\mathbf1_{\{Y<q\}}-\tau$. Difference quotients are bounded by one, so the <dominated convergence theorem> gives
$$
\frac{d}{dq}\mathbb E\rho_\tau(Y-q)=F(q)-\tau.
$$
The continuous, strictly increasing <distribution function> has limits zero and one. Its derivative expression changes sign exactly once, from negative to positive, and hence
$$
\boxed{q_\tau=F^{-1}(\tau).}
$$
Equivalently the difference between the expected losses at $q$ and $q_\tau$ is $\int_{q_\tau}^q(F(t)-\tau)\,dt$, positive whenever $q\ne q_\tau$. This proves <population quantiles minimize check loss> directly.
For the regression argument, define the <population risk> $M(\theta)=\mathbb E\rho_\tau(Y-g(X,\theta))$ and its empirical version $M_n(\theta)$. Boundedness of $g$ and integrability of the error give integrability of $Y$ and of all these losses. Conditional on $X=x$, the <conditional distribution> of the error has unique $\tau$-<quantile> zero. Applying the preceding calculation conditionally shows that its expected loss is uniquely minimized when the fitted displacement $g(x,\theta)-g(x,\theta_0)$ equals zero. Thus
$$
M(\theta)-M(\theta_0)\geq0,
$$
with strict inequality whenever that displacement is nonzero on an event of positive probability. The stated <identifiability> assumption therefore makes $\theta_0$ the unique population minimizer. This is <conditional quantile identification>.
The Lipschitz property of the <check loss> gives
$$
|\rho_\tau(Y-g(X,\theta))-\rho_\tau(Y-g(X,\eta))|\leq |g(X,\theta)-g(X,\eta)|.
$$
Continuity of $g$ and its uniform bound, followed by <dominated convergence>, show that $M$ is continuous on $\Theta$.
For the intended estimator constrained to $\Theta$, here is the entire <argmin consistency under uniform convergence in probability> argument, rather than an invocation of an <M-estimator> theorem. For a fixed $\varepsilon>0$, let $C_\varepsilon=\{\theta\in\Theta:\|\theta-\theta_0\|_2\geq\varepsilon\}$. If this set is empty, there is nothing to prove. Otherwise it is compact, and continuity and uniqueness give a strictly positive gap
$$
d_\varepsilon=\min_{\theta\in C_\varepsilon}(M(\theta)-M(\theta_0))>0.
$$
Let $D_n=\sup_{\theta\in\Theta}|M_n(\theta)-M(\theta)|$. If $\widehat\theta_n\in\Theta$ minimizes $M_n$, then
$$
M(\widehat\theta_n)\leq M_n(\widehat\theta_n)+D_n\leq M_n(\theta_0)+D_n\leq M(\theta_0)+2D_n.
$$
Consequently
$$
\mathbb P(\|\widehat\theta_n-\theta_0\|_2\geq\varepsilon)\leq\mathbb P(2D_n\geq d_\varepsilon)\longrightarrow0.
$$
\b[The constrained estimator is consistent.] Existence of a constrained minimum follows from continuity of the sample criterion and compactness. The argument applies to any measurable choice of minimizer.
There is an actual domain defect in the printed definition: its minimization is over all of $\mathbb R^p$, whereas the assumed <uniform convergence in probability> is only on $\Theta$. \b[That literal unconstrained consistency claim is false.] The preceding conclusion needs minimization over $\Theta$, or another assumption that confines the selected minimizers to a set where the convergence and separation arguments apply. No change of question heading is needed to make this qualification explicit.
Here is a counterexample satisfying even <identifiability> over the whole line. Take a constant covariate, independent standard normal errors, $\tau=1/2$, $\theta_0=0$, $\Theta=[-1,1]$, and
$$
g(x,\theta)=a(\theta)=\frac{\theta^2}{1+\theta^4}.
$$
This is bounded and continuous, has range $[0,1/2]$, and vanishes only at zero. The conditional error <distribution function> is continuous and strictly increasing with median zero. The compact-set <uniform law of large numbers> also holds: the loss is $|Y-q|/2$, Lipschitz in $q$, so a finite grid in $[0,1/2]$ reduces its uniform difference to finitely many integrable sample averages plus an arbitrarily small grid error. The <weak law of large numbers> then gives the asserted convergence.
Let $m_n$ be a <sample median>, and put $c_n=\min(1/2,\max(0,m_n))$. Absolute loss over the range $[0,1/2]$ is minimized at $c_n$. Choose the following global empirical minimizer:
$$
\widehat\theta_n=\begin{cases}
0,&c_n=0,\\
\displaystyle\sqrt{\frac{1+\sqrt{1-4c_n^2}}{2c_n}},&0<c_n\leq1/2.
\end{cases}
$$
Solving the quadratic in $\theta^2$ shows $a(\widehat\theta_n)=c_n$, so this really minimizes the criterion over the whole line. For odd $n$, symmetry and continuity give $\mathbb P(m_n>0)=1/2$. Moreover $m_n\to0$ in probability: for each fixed positive $\delta$, the proportion of observations below $\delta$ converges to a number greater than one half, and the analogous proportion below $-\delta$ converges to a number less than one half. Therefore
$$
\mathbb P(\widehat\theta_n>1)\geq\mathbb P(0<m_n<1/2)\longrightarrow\frac12
$$
along odd $n$. The selected unconstrained estimator is not consistent. This illustrates why a <global argmin may escape a compact convergence set>.
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