Let $\mathcal{B}_1,\mathcal{B}_2$ be some Banach spaces, then $$S=\left\{A:\mathcal{B}_1\to\mathcal{B}_2\big|\left\lVert A\right\rVert\leqslant1\right\}\subset\mathcal{L}\left(\mathcal{B}_1,\mathcal{B}_2\right)$$ is a closed uniball in a linear continuous operator space. Let $x\in\mathcal{B}_1$ be some fixed element. I have a task to find an orbit of such point, which is denoted as: $$\text{Orb}\left(x\right)=Sx=\left\{Ax\big|A\in S\right\}$$
I think, such orbit would be a closed ball of radius $\left\lVert x\right\rVert$ in $\mathcal{B}_2$, but I don't know how to prove it. The furthers I got is that $\text{Orb}\left(x\right)\in\left\{y\in\mathcal{B}_2\big|\left\lVert y\right\rVert\leqslant\left\lVert x\right\rVert\right\}$. What should I do next?
The orbit is indeed the closed ball $\overline{B}(0, \|x\|)$ in $\mathcal{B}_2$.
If $x = 0$ then it is clear. Assume $x \ne 0$.
Let $y \in \overline{B}(0, \|x\|)$ be arbitrary. We will construct a bounded linear map $A : \mathcal{B}_1 \to \mathcal{B}_2$ with $\|A\| \le 1$ such that $Ax = y$.
As @Ashwin Trisal suggests, a corollary of Hahn-Banach implies that there exists a bounded linear functional $f : \mathcal{B}_1 \to \Bbb F$ such that $\|f\| = 1$ and $f(x) = \|x\|$.
Define $A : \mathcal{B}_1 \to \mathcal{B}_2$ as $Av = \frac{f(v)}{\|x\|}y$ for all $v \in \mathcal{B}_1$. Clearly $A$ is linear and we have $Ax = y$. Furthermore $$\|Av\| = \left\|\frac{f(v)}{\|x\|}y\right\| \le \frac{\|v\|\|y\|}{\|x\|}, \forall v \in \mathcal{B}_1 \implies \|A\| \le \frac{\|y\|}{\|x\|} \le 1$$
so $A \in S$. Therefore $y \in \operatorname{Orb}(x)$.
Since $y$ was arbitrary, we conclude $\operatorname{Orb}(x) = \overline{B}(0, \|x\|)$.