We have $S\in\mathbb{B}(\mathcal{H})$ (where $\mathbb{B}(\mathcal{H})$ is algebra of bounded linear operators in Hilbert space) and $S$ is unilateral shift. Compute
$d(S,\mathbb{K(\mathcal{H})})=\inf\{\|S-K\|:K\in \mathbb{K(\mathcal{H})}\}$, where $\mathbb{K(\mathcal{H})}$ are compact operators in Hilbert space.
On
Shift is defined with $S\mu = \sum\limits_{n=1}^\infty \langle\mu, e_n\rangle e_{n+1}$, $\mu \in \mathcal{H}$ and $(e_n)$ is orthonormal basis for $\mathcal{H}$. Also, $\mathcal{H}$ is separable and infinite dimensional Hilbert space.
Since $\lVert S\rVert = 1$, and $0$ is compact, the distance is at most $1$.
Further, the sequence $(e_n)_{n\in\mathbb{Z}^+}$ converges weakly to $0$, hence for every $K \in \mathbb{K}(\mathcal{H})$, we have $K(e_n) \to 0$ strongly. Thus
$$\lVert S-K\rVert \geqslant \sup_{n\in\mathbb{Z}^+} \lVert S(e_n) - K(e_n)\rVert \geqslant \lim_{n\to\infty} \lVert e_{n+1} - K(e_n)\rVert = 1,$$
hence the distance is at least $1$. The distance $1$ is attained for $K = 0$, for example.
(Thanks to Daniel for catching an earlier bug in my proof.)
Let $K$ be a compact operator with finite rank. Since $\mathbb{H}$ is infinite dimensional, we have some $x \neq 0$ such that $Kx=0$. Then $(S-K)x = Sx$, and since we have $\|Sx\|=\|x\|$, we see that $\|S-K\| \ge 1$.
Now let $K$ be any compact operator. Then we have compact, finite rank operators $K_n$ such that $K_n \to K$. Since $\|S-K_n\| \ge 1$, and $\|S-K_n\| \le \|S-K\|+\|K-K_n\|$, we see that $\|S-K\| \ge 1$.
The operator $K=0$ is compact, and $\|S-K\| = \|S\| = 1$, hence $K$ is a minimizer. Hence the distance from $S$ to the set of compact operators is one.
Note: The 'only' requirement on $S$ is that it be bounded and $\|Sx\| = \|x\|$ for all $x$. The proof works in any Banach space.