I'm checking if below result is true by giving a proof. Could you verify if my proof is fine?
Let $G, L$ be subsets of a normed linear space $(E, |\cdot|)$. Let $f:E \to \mathbb R$ be linear and $G+L:=\{x+y\mid x\in G, y\in L\}$. Let $\alpha:=\sup_{x\in G} f(x), \beta:= \sup_{y\in L} f(y)$, and $\gamma:= \sup_{z\in G+L} f(z)$. Assume that $\alpha, \beta, \gamma \in \mathbb R$. Then $\alpha +\beta = \gamma$.
My attempt: Let $(x_n)$ and $(y_n)$ be sequences in $G$ and $L$ respectively such that $f(x_n) \to \alpha$ and $f(y_n) \to \beta$. Such sequences exist by axiom of choice (AC). Then $$\begin{aligned} \gamma &\ge \lim_{n \to \infty} f(x_n+y_n)\\ &= \lim_{n \to \infty} \big (f(x_n) + f(y_n) \big )\\ &= \lim_{n \to \infty} f(x_n) + \lim_{n \to \infty} f(y_n) \\ & = \alpha + \beta. \end{aligned}$$
By AC, there is a sequence $(z_n)$ in $G+L$ such that $f(z_n) \to \gamma$. By AC again, there exist sequences $(x_n)$ in $G$ and $(y_n)$ in $L$ such that $z_n = x_n + y_n$. Then $$\begin{aligned} \gamma &= \lim_{n \to \infty} f(z_n) = \limsup_{n \to \infty} f(z_n) \\ &= \limsup_{n \to \infty} \big (f(x_n)+f(y_n)\big) \\ &\le \limsup_{n \to \infty} f(x_n) + \limsup_{n \to \infty} f(y_n) \\ &=\alpha + \beta. \end{aligned}$$
This completes the proof.