[proofplan]
We choose one feasible point to obtain a nonempty bounded sublevel set that contains every near-minimizer. Closedness of $C$ and lower semicontinuity of $f$ make this sublevel set closed, hence compact in finite-dimensional Euclidean space. A minimizing sequence therefore has a convergent subsequence, and lower semicontinuity identifies the limit as an actual minimizer. Finally, the solution set is a closed subset of the same compact sublevel set, so it is compact.
[/proofplan]
[step:Fix the optimal value and place near-minimizers in one bounded sublevel set]
Define the optimal value $p^* \in \mathbb{R}$ by
\begin{align*}
p^* := \inf_{x \in C} f(x).
\end{align*}
Define the solution set $S \subset C$ by
\begin{align*}
S := \{x \in C : f(x) = p^*\}.
\end{align*}
This number is finite because the problem is bounded below and feasible. Indeed, boundedness below gives $p^* > -\infty$, while feasibility gives a point $x_0 \in C \cap \operatorname{dom} f$, so $f(x_0) \in \mathbb{R}$ and hence $p^* \leq f(x_0) < +\infty$.
Fix such a point $x_0 \in C \cap \operatorname{dom} f$ and define $\alpha := f(x_0) \in \mathbb{R}$. Define $\beta := \alpha + 1 \in \mathbb{R}$. The sublevel set
\begin{align*}
L_\beta := \{x \in C : f(x) \leq \beta\}
\end{align*}
is nonempty because $x_0 \in L_\beta$, and it is bounded by level-boundedness on $C$.
[/step]
[step:Show the relevant sublevel set is compact]
The set $\{x \in \mathbb{R}^n : f(x) \leq \beta\}$ is closed in $\mathbb{R}^n$ because $f$ is lower semicontinuous. Since $C$ is closed in $\mathbb{R}^n$, the intersection
\begin{align*}
L_\beta = C \cap \{x \in \mathbb{R}^n : f(x) \leq \beta\}
\end{align*}
is closed in $\mathbb{R}^n$. We already know that $L_\beta$ is bounded, so the Heine-Borel finite-dimensional compactness criterion for subsets of $\mathbb{R}^n$ implies that $L_\beta$ is compact.
[guided]
The goal of this step is to turn level-boundedness into compactness. Level-boundedness gives only boundedness, so we must also prove closedness.
First, lower semicontinuity of $f: \mathbb{R}^n \to (-\infty,+\infty]$ means that every sublevel set of the form
\begin{align*}
\{x \in \mathbb{R}^n : f(x) \leq \gamma\}
\end{align*}
is closed in $\mathbb{R}^n$ for each $\gamma \in \mathbb{R}$. Applying this with $\gamma = \beta$, the set
\begin{align*}
\{x \in \mathbb{R}^n : f(x) \leq \beta\}
\end{align*}
is closed. The constraint set $C$ is closed by hypothesis. Therefore their intersection
\begin{align*}
L_\beta = C \cap \{x \in \mathbb{R}^n : f(x) \leq \beta\}
\end{align*}
is closed in $\mathbb{R}^n$.
Second, level-boundedness on $C$ says precisely that $L_\beta$ is bounded. Thus $L_\beta$ is both closed and bounded in the finite-dimensional Euclidean space $\mathbb{R}^n$. By the Heine-Borel finite-dimensional compactness criterion, closed and bounded subsets of $\mathbb{R}^n$ are compact. Hence $L_\beta$ is compact.
[/guided]
[/step]
[step:Extract a convergent subsequence from a minimizing sequence]
Let $\mathbb{N} := \{1,2,3,\dots\}$ denote the set of positive integers. For each $k \in \mathbb{N}$, the definition of the infimum gives a point $x_k \in C$ such that
\begin{align*}
f(x_k) \leq p^* + \frac{1}{k}.
\end{align*}
Since $p^* \leq \alpha$ and $\beta = \alpha + 1$, we have
\begin{align*}
f(x_k) \leq p^* + \frac{1}{k} \leq \alpha + 1 = \beta
\end{align*}
for every $k \in \mathbb{N}$. Hence $x_k \in L_\beta$ for every $k \in \mathbb{N}$. Also,
\begin{align*}
p^* \leq f(x_k) \leq p^* + \frac{1}{k},
\end{align*}
so $f(x_k) \to p^*$ as $k \to \infty$.
Since $L_\beta$ is compact and $(x_k)_{k \in \mathbb{N}}$ is a sequence in $L_\beta$, there exist a subsequence $(x_{k_j})_{j \in \mathbb{N}}$ and a point $\bar{x} \in L_\beta$ such that $x_{k_j} \to \bar{x}$ in $\mathbb{R}^n$.
[/step]
[step:Use lower semicontinuity to prove the limit is a minimizer]
Since $f$ is lower semicontinuous and $x_{k_j} \to \bar{x}$, we have
\begin{align*}
f(\bar{x}) \leq \liminf_{j \to \infty} f(x_{k_j}).
\end{align*}
The full sequence $f(x_k)$ converges to $p^*$, hence the subsequence also satisfies
\begin{align*}
\liminf_{j \to \infty} f(x_{k_j}) = p^*.
\end{align*}
Therefore $f(\bar{x}) \leq p^*$. Since $\bar{x} \in L_\beta \subset C$ and $p^*$ is the infimum of $f$ over $C$, we also have $p^* \leq f(\bar{x})$. Hence
\begin{align*}
f(\bar{x}) = p^*.
\end{align*}
Thus $\bar{x} \in S$, so the solution set $S$ is nonempty.
[/step]
[step:Identify the solution set as a closed subset of a compact set]
Because $p^* \leq \beta$, every minimizer belongs to $L_\beta$. Hence
\begin{align*}
S \subset L_\beta.
\end{align*}
Moreover, since $p^*$ is a lower bound for $f$ on $C$,
\begin{align*}
S = C \cap \{x \in \mathbb{R}^n : f(x) \leq p^*\}.
\end{align*}
The set $\{x \in \mathbb{R}^n : f(x) \leq p^*\}$ is closed by lower semicontinuity of $f$, and $C$ is closed by hypothesis. Therefore $S$ is closed in $\mathbb{R}^n$. Since $S$ is a closed subset of the compact set $L_\beta$, it is compact. This proves that the solution set is nonempty and compact.
[/step]
