I'm reading the proof of Fenchel-Rockafellar's theorem.
Let $E$ be a topological vector space and $\varphi, \psi: E \to \mathbb R \cup \{+\infty\}$ convex functions. Assume there exists $x_{0} \in E$ such that $\varphi(x_0) < +\infty, \psi(x_0) < +\infty$, and $\varphi$ is continuous at $x_{0}$. Then $$ \begin{aligned} \inf _{x \in E}\{\varphi(x)+\psi(x)\} &=\sup _{f \in E^{\star}}\left\{-\varphi^{\star}(-f)-\psi^{\star}(f)\right\} \\ &=\max _{f \in E^{\star}}\left\{-\varphi^{\star}(-f)-\psi^{\star}(f)\right\}. \end{aligned} $$
In the proof, the author defines $C := \operatorname{epi} \varphi$ and claims that $\operatorname{int} C \neq \emptyset$ because $\varphi$ is continuous at $x_0$. It seems to me it's enough that $\varphi$ is upper semi-continuous at $x_0$.
Let $\varepsilon > 0$ and $\lambda_0 := \varphi (x_0) + 2\varepsilon$. Then $(x_0, \lambda_0) \in C$. Because $\varphi$ is upper semi-continuous at $x_0$, there is a neighborhood $U$ of $x_0$ such that $\varphi(x) < \varphi (x_0) + \varepsilon$ for all $x \in U$. Let $I := (\varphi (x_0) + \varepsilon, \varphi (x_0) + 3\varepsilon)$. Then $I$ is a neighborhood of $\lambda_0$. Clearly, $\varphi(x) < \lambda$ for all $(x, \lambda) \in U \times I$, i.e., $U \times I \subseteq C$. It follows that $(x_0, \lambda_0) \in \operatorname{int} C$.
Could you confirm if my understanding is correct?