No, an optimal solution need not exist. Take $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = e^{x}$.

However if you restrict $S$ to be compact instead of just closed, then you are guaranteed a solution. For a simple proof, let $f(x_n) \to \inf_{x \in S} f(x)$. $x_n \in S$ has a convergent subsequence by compactness, and let $x \in S$ be its limit point. By lower semicontinuity $x$ minimizes $f$ on $S$. Lastly, note that the infimum above must be finite because $f$ is proper, hence $f(x) \neq \pm \infty$.

No, an optimal solution need not exist. Take $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = e^{x}$.

However if you restrict $S$ to be compact instead of just closed, then you are guaranteed a solution. In fact, convexity is not required. For a simple proof, let $f(x_n) \to \inf_{x \in S} f(x)$. $x_n \in S$ has a convergent subsequence by compactness, and let $x \in S$ be its limit point. By lower semicontinuity $x$ minimizes $f$ on $S$. Lastly, note that the infimum above must be finite because $f$ is proper, hence $f(x) \neq \pm \infty$.