I'm trying to prove this lemma. Could you verify if my attempt is fine?
Let $X$ be a topological vector space and $C$ a convex subset of $X$. Then $\operatorname{int} C$ and $\overline C$ are convex.
My attempt: Let $x, y \in \operatorname{int} C$. There are open sets $C_x, C_y \subseteq C$ such that $x \in C_x$ and $y \in C_y$. Let $$D := \bigcup \left \{\lambda u +(1-\lambda )C_y \mid (\lambda , u) \in [0,1] \times C_x \right \}.$$
We have $\lambda u +(1-\lambda )C_y$ is open for all $(\lambda , u) \in [0,1] \times C_x$, so $D$ is open. Also, $\lambda x + (1-\lambda )y \in D$ for all $\lambda \in [0, 1]$. Moreover, $D \subseteq C$ because $C$ is convex. Hence $\lambda x + (1-\lambda )y \in \operatorname{int} C$ for all $\lambda \in [0, 1]$. Hence $\operatorname{int} C$ is convex.
Let $x, y \in \overline C$ and $\lambda \in [0, 1]$. Let $z := \lambda x + (1-\lambda) y$. We define a map $f: X \times X \to X$ by $f(u, v) = \lambda u + (1-\lambda) v$. Let $U$ be an open neighborhood of $z$. To prove that $z \in \overline C$, we'll show that $U \cap C \neq \emptyset$. Because $f$ is continuous, there are open neighborhoods $V_x$ of $x$ an $V_y$ of $y$ such that $f(V_x \times V_y) \subseteq U$. Because $x, y \in \overline C$, there are $x' \in V_x \cap C$ and $y' \in V_y \cap C$. Also, $f(x', y') = \lambda x' + (1-\lambda) y' \in U$. Since $C$ is convex $\lambda x' + (1-\lambda) y' \in C$. Hence $\lambda x' + (1-\lambda) y' \in U \cap C$.