Consider the vector space $C([-\pi,\pi])$ of all continuous functions on the interval $[-\pi,\pi]$ equipped with the norm $$||f||= \max_{x\in[-\pi,\pi]}\{|f(x)|\}.$$
sketch the closed ball of center $\sin x$ and radius one.
$\textbf{Attempt:}$ I'm not sure what this question really is asking. Wouldn't the closed ball simply be two the maxima of $|\sin x|$ in the interval $[-\pi,\pi]$. Meaning just the point 1?
On
The closed ball $\bar{B}(\sin,1)$ consists of continuous $f$ such that $|f(x)-\sin x| \le 1$ for all $x$.
To visualise, form the curves $x \mapsto \sin x \pm 1$, then the closed ball is all the continuous functions whose graph lies inside the 'envelope' formed by these curves.
Or, imagine putting a pipe of radius one around the sine curve, then all continuous $f$ that lie in or on the pipe form the closed ball around $\sin$.
The basic idea is that you need to resort to the definition of a closed ball.
In a metric space with distance function $d(x, y)$, the closed ball centered at $x_0$ with radius $r$ is given by
$$\overline{B_r}(x_0) = \{x : d(x, x_0) \leq r\}$$
(note that in my comment to the question, I had mistakenly used "$=$", when I meant "$\leq$"; it's all about the definitions!). In a normed space with norm $\lVert \cdot \rVert$, it's typical to define the distance function
$$d(x, y) \overset{\text{def}}{=} \lVert x - y \rVert,$$
so that the definition of closed ball above becomes $$\overline{B_r}(x_0) = \{x : \lVert x - x_0 \rVert \leq r\}.$$
Your job is to rewrite this, using your norm on the function space $C([-\pi, \pi])$ for your particular example. Then you can start to think about what functions are in your closed ball.
As far as sketching goes, it's not very easy to sketch anything besides functions in any function space. In this case, you can sketch the region of $\Bbb R^2$ in which the graphs of functions in $\overline{B_1}\big(\sin(x)\big)$ have to live.