How did Behrend obtain $R_\lambda(x) = \text{co}\{-\lambda x,+\lambda x\}$ in his $M$-structure lecture notes?

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Currently I am reading M-Stuture and Banach-Stone Theorem by Behrend, page $55,$ section multiplier. I encounter the following terminology.

For each $x\in X$ and $\lambda>0,$ denote $$R_\lambda(x) = \bigcap\bigg\{ D\,\,|\,\,D\text{ is a closed ball such that }\{\mu x:\mu\in\mathbb{R},|\mu|\leq\lambda\}\subseteq D \bigg\}.$$

Behrend provided an example to determine $R_\lambda(x)$ at page $56,$ which goes as follows:

For example, consider the real space $\mathbb{R}^2$ together with the norms $\|(a,b)\| = (a^2+b^2)^{\frac{1}{2}}.$ Then $R_\lambda(x) = \text{co}\{-\lambda x,+\lambda x\}$ for every $\lambda>0$ and $x\in X.$

Note that co$\{-\lambda x,+\lambda x\}$ is the convex hull of the two points.

However, I do not 'see' how is this the case. In particular,

Question: For each $x\in X$ and $\lambda>0,$ how to show that $$R_\lambda(x) = \text{co}\{-\lambda x,+\lambda x\}?$$

Any hint is appreciated.

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The containment $\supseteq$ is obvious from the definition of $R_\lambda$.
For the converse, in case $X=\Bbb R^2$, consider any point $p$ outside of that segment $S=\mathrm{co}(-\lambda x, \lambda x) $. Then there is a (perhaps big) closed disk $D$ with chord $S$ that avoids $p$.