The weak topology is compatible with the vector space structure I'm trying to prove this result

Let $E$ be a topological vector space and $E^\star$ its topological dual. Let $\sigma(E, E^\star)$ be the weak topology on $E$. We denote by $E_w$ the vector space $E$ together with $\sigma(E, E^\star)$. Then $E_w$ is a topological vector space.

I posted my proof as below answer. Could you have a check on my attempt?

First, we consider the linear map $T: E_w \times E_w \to E_w, (x, y) \mapsto x+y$. We want to prove that $T$ is continuous. It suffices to show that $$f \circ T: E_w \times E_w \to \mathbb R, (x, y) \mapsto f(x)+f(y).$$ is continuous for all $f \in E^\star$. The claim then follows from below diagram
$$
\substack{E_w \times E_w \\ (x,y)}  \, \substack{ \longrightarrow \\ \longmapsto } \, \substack{\mathbb R \times \mathbb R \\ (f(x),f(y))} \, \substack{ \longrightarrow \\ \longmapsto } \, \substack{ \mathbb R \\ f(x)+f(y)}
$$
and the fact that $f$ is also continuous in the weak topology. Second, we consider the linear map $L: \mathbb R \times E_w \to E_w, (t, x) \mapsto tx$. We want to prove that $L$ is continuous. It suffices to show that $$f \circ L: \mathbb R \times E_w \to \mathbb R, (t, x) \mapsto tf(x).$$ is continuous for all $f \in E^\star$. The claim then follows from below diagram
$$
\substack{\mathbb R \times E_w \\ (t, x)}  \, \substack{ \longrightarrow \\ \longmapsto } \, \substack{\mathbb R \times \mathbb R \\ (t,f(x))} \, \substack{ \longrightarrow \\ \longmapsto } \, \substack{ \mathbb R \\ tf(x)}
$$
and the fact that $f$ is also continuous in the weak topology.
