[proofplan] We prove both implications directly from the definitions. If $f$ is lower semicontinuous, then any point strictly below the epigraph has a product neighbourhood still lying below the epigraph, so the complement of the epigraph is open. Conversely, if the epigraph is closed, then for a convergent sequence $x_k \to x_0$ we compare $f(x_0)$ with $L := \liminf_{k \to \infty} f(x_k)$ by placing suitable heights above the graph and passing to limits in the closed epigraph. The exceptional case $L=-\infty$ is ruled out by the fact that $f$ never takes the value $-\infty$. [/proofplan]

[step:Show that lower semicontinuity makes the complement of the epigraph open]Assume $f$ is lower semicontinuous on $\mathbb{R}^n$. We prove that $\mathbb{R}^{n+1} \setminus \operatorname{epi} f$ is open. Let $(x_0,r_0) \in \mathbb{R}^{n+1} \setminus \operatorname{epi} f$. Then $r_0 < f(x_0)$. Choose a real number $a \in \mathbb{R}$ such that \begin{align*} r_0 < a < f(x_0). \end{align*} If $f(x_0)=+\infty$, take for instance $a=r_0+1$. By lower semicontinuity at $x_0$, the strict sublevel complement \begin{align*} \{x \in \mathbb{R}^n : f(x) > a\} \end{align*} is a neighbourhood of $x_0$. Hence there exists $\delta_1>0$ such that $|x-x_0|<\delta_1$ implies $f(x)>a$. Define $\delta_2 := a-r_0>0$ and $\delta := \min\{\delta_1,\delta_2\}>0$. If $(x,r) \in \mathbb{R}^{n+1}$ satisfies $|x-x_0|<\delta$ and $|r-r_0|<\delta$, then \begin{align*} r < r_0+\delta \le r_0+\delta_2 = a < f(x). \end{align*} Thus $(x,r) \notin \operatorname{epi} f$. Therefore a Euclidean neighbourhood of $(x_0,r_0)$ is contained in $\mathbb{R}^{n+1}\setminus \operatorname{epi} f$, so the complement is open. Hence $\operatorname{epi} f$ is closed.[/step]

[guided]Assume $f$ is lower semicontinuous on $\mathbb{R}^n$. To prove that $\operatorname{epi} f$ is closed, it is enough to prove that its complement is open. Take a point $(x_0,r_0) \notin \operatorname{epi} f$. By the definition of the epigraph, this means \begin{align*} r_0 < f(x_0). \end{align*} We need to build a full neighbourhood of $(x_0,r_0)$ that remains outside the epigraph. The gap between $r_0$ and $f(x_0)$ is what gives room to do this. Choose $a \in \mathbb{R}$ with \begin{align*} r_0 < a < f(x_0). \end{align*} If $f(x_0)=+\infty$, the choice $a=r_0+1$ works. Lower semicontinuity at $x_0$ says that strict lower bounds persist locally: since $a<f(x_0)$, there exists $\delta_1>0$ such that \begin{align*} |x-x_0|<\delta_1 \implies f(x)>a. \end{align*} Now define $\delta_2 := a-r_0>0$ and $\delta := \min\{\delta_1,\delta_2\}$. If $(x,r)$ satisfies $|x-x_0|<\delta$ and $|r-r_0|<\delta$, then the vertical coordinate satisfies \begin{align*} r < r_0+\delta \le r_0+\delta_2 = a. \end{align*} At the same time, the horizontal coordinate satisfies $|x-x_0|<\delta \le \delta_1$, so $f(x)>a$. Combining these inequalities gives \begin{align*} r < a < f(x). \end{align*} Thus $f(x)\le r$ is false, so $(x,r)\notin \operatorname{epi} f$. We have found a neighbourhood of $(x_0,r_0)$ contained in the complement of the epigraph. Therefore the complement is open, and $\operatorname{epi} f$ is closed.[/guided]

[step:Use closedness of the epigraph to prove the lower semicontinuity inequality] Assume $\operatorname{epi} f$ is closed in $\mathbb{R}^{n+1}$. Let $(x_k)_{k=1}^{\infty}: \mathbb{N}\to\mathbb{R}^n$ be a sequence such that $x_k \to x_0$ in $\mathbb{R}^n$. Define \begin{align*} L := \liminf_{k \to \infty} f(x_k) \in [-\infty,+\infty]. \end{align*} We prove \begin{align*} f(x_0) \le L. \end{align*} If $L=+\infty$, then $f(x_0)\le+\infty=L$, so the desired inequality holds. Suppose next that $L\in\mathbb{R}$. Fix $\varepsilon>0$. By the definition of the limit inferior, there is a subsequence $(x_{k_j})_{j=1}^{\infty}$ such that \begin{align*} f(x_{k_j}) \to L. \end{align*} After discarding finitely many terms, we may assume $f(x_{k_j})\le L+\varepsilon$ for every $j$. For each $j$, define $r_j := L+\varepsilon$. Then $(x_{k_j},r_j)\in\operatorname{epi} f$ for every $j$, and \begin{align*} (x_{k_j},r_j) \to (x_0,L+\varepsilon) \end{align*} in $\mathbb{R}^{n+1}$. Since $\operatorname{epi} f$ is closed, $(x_0,L+\varepsilon)\in\operatorname{epi} f$. Hence \begin{align*} f(x_0)\le L+\varepsilon. \end{align*} Because $\varepsilon>0$ was arbitrary, we obtain $f(x_0)\le L$. It remains to rule out $L=-\infty$. If $L=-\infty$, then for every $a\in\mathbb{R}$ and every $m\in\mathbb{N}$ there exists $k\ge m$ such that $f(x_k)\le a$. Thus we may choose a subsequence $(x_{k_j})_{j=1}^{\infty}$ with $f(x_{k_j})\le a$ for every $j$. Then $(x_{k_j},a)\in\operatorname{epi} f$ for every $j$, and \begin{align*} (x_{k_j},a) \to (x_0,a) \end{align*} in $\mathbb{R}^{n+1}$. Closedness gives $(x_0,a)\in\operatorname{epi} f$, so $f(x_0)\le a$. Since $a\in\mathbb{R}$ was arbitrary, this implies $f(x_0)\le a$ for all real $a$, which is impossible because $f(x_0)\in(-\infty,+\infty]$. Therefore $L=-\infty$ cannot occur. Thus for every sequence $x_k\to x_0$, \begin{align*} f(x_0)\le \liminf_{k\to\infty} f(x_k). \end{align*} This is precisely lower semicontinuity of $f$ at $x_0$. Since $x_0\in\mathbb{R}^n$ was arbitrary, $f$ is lower semicontinuous on $\mathbb{R}^n$. [/step]