Question: Show that the distance $d(x)$ of a point $x = (x_k)\in \ell_\infty$ to $c_0$ is equal to $\limsup_{k\to\infty} |x_k|.$ Thus the norm in $\ell_\infty / c_0$ is $\|\hat{x}\| = \limsup_{k\to\infty} |x_k|.$
My attempt:
Note that $$ \|x+c_0\| = \inf_{y\in c_0}\sup_{k\geq 1} |x_k-y_k| \quad\text{and}\quad \limsup_{k\to\infty} |x_k| = \inf_{n\geq 1} \sup_{k\geq n}|x_k| . $$ Pick any $n\geq 1.$ Let $y = (y_k)_{k=1}^\infty$ be such that $y_k = x_k$ for all $k\leq n$ and $y_k = 0$ for all $k\geq n+1.$ Clearly $\lim_{k\to\infty} y_k = 0$ and so $y\in c_0.$ It follows that $$ \sup_{k\geq n}|x_k| = \sup_{k\geq 1}|x_k-y_k| \geq \inf_{y\in c_0} \sup_{k\geq 1} |x_k-y_k|. $$ Since $n\geq 1$ is arbitrary, we conclude that $$ \inf_{y\in c_0}\sup_{k\geq 1} |x_k-y_k|\leq \inf_{n\geq 1} \sup_{k\geq n}|x_k| . $$
On the other hand, pick any $y\in c_0.$ Let $\epsilon>0$ be given. Then there exists $N\in\mathbb{N}$ such that for all $k\geq N,$ we have $|y_k|<\epsilon.$ It follows that $$ \sup_{k\geq 1} |x_k-y_k| \geq \sup_{k\geq N}|x_k-y_k| \geq \sup_{k\geq N} \left( |x_k| - |y_k| \right) \geq \sup_{k\geq N}\left( |x_k| - \epsilon \right) = \sup_{k\geq N} |x_k| - \epsilon \geq \inf_{n\geq 0} \sup_{k\geq n}|x_k| - \epsilon. $$ Since $\epsilon>0$ is arbitrary, we have $$ \sup_{k\geq 1}|x_k-y_k| \geq \inf_{n\geq 1}\sup_{k\geq n} |x_k|. $$ Since $y\in c_0$ is also arbitrary, we conclude that $$ \inf_{y\in c_0}\sup_{k\geq 1}|x_k-y_k| \geq \inf_{n\geq 1}\sup_{k\geq n} |x_k|. $$
Is my proof above correct?