I'm doing Exercise 1.7 from textbook Functional Analysis, Sobolev Spaces and Partial Differential Equations. Could you verify if my attempt is fine?
Let $X$ be a topological vector space and $C$ a convex subset of $X$. If $\operatorname{int} C \neq \emptyset$, then $\overline C = \overline{\operatorname{int} C}$.
My attempt: Let $x \in C$, $y \in \operatorname{int} C$, and $\lambda \in [0, 1)$. Let $z := \lambda x + (1-\lambda)y$. There is a neighborhood $V$ of $y$ such that $V \subseteq C$. Let $D := \lambda x + (1-\lambda) V$. Because $\lambda \neq 1$, $D$ is a neighborhood of $z$. Because $C$ is convex, $D \subseteq C$. Hence $z \in \operatorname{int} C$.
It's clear that $\overline{\operatorname{int} C} \subseteq \overline C$. Let's prove the reverse inclusion, i.e., $\overline C \subseteq \overline{\operatorname{int} C}$. Let $x \in \overline C$ and $V$ be a neighborhood of $x$. Our goal is to show that $V \cap {\operatorname{int} C} \neq \emptyset$. First, there is $y \in V \cap C$. Because $\operatorname{int} C \neq \emptyset$, there is $z \in \operatorname{int} C$. Consider the map $$f:\mathbb R \to X, \quad \lambda \mapsto \lambda z + (1-\lambda) y.$$
Then $f$ is continuous by definition of a topological vector space. Consider the nets $(\lambda_n)$ and $(y_n)$ where $\lambda_n := 1/(n+1)$ and $y_n := f(\lambda_n)$. We have $\lambda_n \to 0$, so $y_n \to y$. By definition of net convergence, there is $N \in \mathbb N$ such that $y_n \in V$ for all $n \ge N$. As proved above, $\lambda_n \in \operatorname{int} C$ for all $n$. Hence $\lambda_n \in V \cap \operatorname{int} C$ for all $n \ge N$. This completes the proof.