**Existence.** The map $\theta \mapsto \bar{\ell}_n(\theta) = \frac{1}{n}\sum_{i=1}^n \log f(X_i, \theta)$ is continuous on the compact set $\Theta$ (by condition 3 and finiteness from condition 6). By the extreme value theorem, it attains its maximum, so an MLE exists. **Consistency.** Fix $\varepsilon > 0$ and define $\Theta_\varepsilon = \{\theta \in \Theta : \|\theta - \theta_0\| \geq \varepsilon\}$. This is compact (intersection of the compact set $\Theta$ with the closed set $\{\|\theta - \theta_0\| \geq \varepsilon\}$). Since $\ell$ is continuous and $\theta_0 \notin \Theta_\varepsilon$ is the unique maximizer of $\ell$ on $\Theta$, the maximum of $\ell$ on $\Theta_\varepsilon$ satisfies \begin{align*} c(\varepsilon) := \sup_{\theta \in \Theta_\varepsilon} \ell(\theta) < \ell(\theta_0). \end{align*} Choose $\delta(\varepsilon) > 0$ with $c(\varepsilon) + \delta(\varepsilon) < \ell(\theta_0) - \delta(\varepsilon)$. Define the event \begin{align*} A_n(\varepsilon) = \left\{ \sup_{\theta \in \Theta} |\bar{\ell}_n(\theta) - \ell(\theta)| < \delta(\varepsilon) \right\}. \end{align*} By the uniform law of large numbers, $\mathbb{P}(A_n(\varepsilon)) \to 1$. On $A_n(\varepsilon)$, we bound $\sup_{\Theta_\varepsilon} \bar{\ell}_n$ using the triangle inequality: \begin{align*} \sup_{\theta \in \Theta_\varepsilon} \bar{\ell}_n(\theta) \leq \sup_{\theta \in \Theta_\varepsilon} \ell(\theta) + \sup_{\theta \in \Theta} |\bar{\ell}_n(\theta) - \ell(\theta)| < c(\varepsilon) + \delta(\varepsilon). \end{align*} On $A_n(\varepsilon)$ we also have $\bar{\ell}_n(\theta_0) > \ell(\theta_0) - \delta(\varepsilon)$. By choice of $\delta(\varepsilon)$, \begin{align*} \sup_{\theta \in \Theta_\varepsilon} \bar{\ell}_n(\theta) < c(\varepsilon) + \delta(\varepsilon) < \ell(\theta_0) - \delta(\varepsilon) < \bar{\ell}_n(\theta_0). \end{align*} If $\hat{\theta}_n \in \Theta_\varepsilon$, then $\bar{\ell}_n(\hat{\theta}_n) \leq \sup_{\Theta_\varepsilon} \bar{\ell}_n < \bar{\ell}_n(\theta_0)$, contradicting the definition of the MLE. Therefore on $A_n(\varepsilon)$, $\hat{\theta}_n \notin \Theta_\varepsilon$, i.e., $A_n(\varepsilon) \subseteq \{\|\hat{\theta}_n - \theta_0\| < \varepsilon\}$. Since $\mathbb{P}(A_n(\varepsilon)) \to 1$, we conclude $\mathbb{P}(\|\hat{\theta}_n - \theta_0\| < \varepsilon) \to 1$.