[example: The Shift That Refuses Compactness]
Let $H=\ell^2$ and define the right shift $S:H\to H$ by
\begin{align*}
S(a_1,a_2,a_3,\ldots)=(0,a_1,a_2,a_3,\ldots).
\end{align*}
For $x=(a_k)_{k=1}^\infty$ and $y=(b_k)_{k=1}^\infty$ in $\ell^2$, and scalars $\alpha,\beta$, we have
\begin{align*}
S(\alpha x+\beta y)
&=S(\alpha a_1+\beta b_1,\alpha a_2+\beta b_2,\ldots)\\
&=(0,\alpha a_1+\beta b_1,\alpha a_2+\beta b_2,\ldots)\\
&=\alpha(0,a_1,a_2,\ldots)+\beta(0,b_1,b_2,\ldots)\\
&=\alpha Sx+\beta Sy,
\end{align*}
so $S$ is linear. Also,
\begin{align*}
\|Sx\|_{\ell^2}^2
&=|0|^2+|a_1|^2+|a_2|^2+|a_3|^2+\cdots\\
&=\sum_{k=1}^\infty |a_k|^2\\
&=\|x\|_{\ell^2}^2,
\end{align*}
so $\|Sx\|_{\ell^2}=\|x\|_{\ell^2}$ for every $x\in \ell^2$, and $S$ is bounded.
Let $e_k$ be the sequence with $1$ in the $k$-th coordinate and $0$ elsewhere. Then
\begin{align*}
\|e_k\|_{\ell^2}^2=\sum_{n=1}^\infty |(e_k)_n|^2=1,
\end{align*}
so $(e_k)_{k=1}^\infty$ is bounded. Moreover,
\begin{align*}
Se_k=e_{k+1},
\end{align*}
because the single nonzero coordinate of $e_k$ is shifted from position $k$ to position $k+1$.
For $j\ne k$, the sequence $e_j-e_k$ has coordinate $1$ in position $j$, coordinate $-1$ in position $k$, and $0$ in every other position. Hence
\begin{align*}
\|e_j-e_k\|_{\ell^2}^2
&=\sum_{n=1}^\infty |(e_j-e_k)_n|^2\\
&=|1|^2+|-1|^2\\
&=2.
\end{align*}
Therefore, for $j\ne k$,
\begin{align*}
\|Se_j-Se_k\|_{\ell^2}
&=\|e_{j+1}-e_{k+1}\|_{\ell^2}\\
&=\sqrt{2}.
\end{align*}
Every subsequence of $(Se_k)_{k=1}^\infty$ still has pairwise distances $\sqrt{2}$ between distinct terms, so no subsequence is Cauchy. Since every norm-convergent sequence is Cauchy, no subsequence of $(Se_k)_{k=1}^\infty$ converges in norm.
Thus $S$ sends the bounded set $\{e_k:k\in\mathbb{N}\}$ to a set containing a sequence with no norm-convergent subsequence. Its image is not relatively compact, so the bounded linear right shift is not compact.
[/example]