I'm working on a proof that if $\Omega$ is a convex subset of a topological vector space $X$, then so is its closure, $\overline{\Omega}$. My proof was different than the proof in the book, so I want to verify that it is correct. Here it is:
Let $x,y\in\overline{\Omega}$ and $0<\lambda<1$. Suppose $V$ is a neighborhood of the origin, and choose a balanced neighborhood $U$ of the origin such that $U+U\subset V$. Then since $U+x$ and $U+y$ are neighborhoods of $x$ and $y$ respectively, $(U+x)\cap \Omega\neq\emptyset$ and $(U+y)\cap \Omega\neq\emptyset$, so let $a\in (U+x)\cap\Omega$ and $b\in (U+y)\cap \Omega$. Then since $\Omega$ is convex, $\lambda a+(1-\lambda)b\in\Omega$. Additionally, since $U$ is balanced, we have that \begin{align*} \lambda a+(1-\lambda)b&\in \lambda(U+x)+(1-\lambda)(U+y)\\ &=\lambda U + (1-\lambda)U+\lambda x + (1-\lambda)y\\ &\subset U + U + \lambda x + (1-\lambda)y\\ &\subset V+\lambda x + (1-\lambda)y, \end{align*} so that $(V+\lambda x+(1-\lambda)y)\cap \Omega\neq\emptyset$. Therefore, $\lambda x+(1-\lambda)y\in\overline{\Omega}$. This shows $\overline{\Omega}$ is convex.
Was my proof correct? If not, what should I fix? Thank you for your help!