Compact operator without eigenvalues?

Note that $$T^2 = \begin{pmatrix}lMrM&0\\0&rMlM\end{pmatrix},$$ and hence the eigenvectors of $T^2$ are $$v_j = (e_j, 0) , \qquad w_j = (0, e_j),$$ with corresponding eigenvalues $$\lambda_j = \frac{1}{(1 + |j|) (1 + |j+1|)} \, , \qquad \mu_j = \frac{1}{(1 + |j|) (1 + |j-1|)} \, ,$$ respectively. In particular, the eigenspaces of $T^2$ are four-dimensional: for $j \ge 0$, the eigenspace corresponding to $\lambda_j = \mu_{j+1} = \lambda_{-j-1} = \mu_{-j}$ is spanned by $v_j$, $u_{j+1}$, $v_{-j-1}$ and $u_{-j}$.

If $u$ is an eigenvector of $T$, then it is also an eigenvector of $T^2$, and hence it is a linear combination of $v_j$, $u_{j+1}$, $v_{-j-1}$ and $u_{-j}$ for some $j \ge 0$. By inspection, $$ T(a v_j + b u_{j+1} + c v_{-j-1} + d u_{-j}) = \frac{a u_{j+1} + d v_{-j-1}}{1 + |j|} + \frac{b v_j + c u_{-j}}{1 + |j + 1|} $$ corresponds to a simple block $4\times4$ matrix, and it is now an elementary exercise to find the eigenvectors.

Sure, let $\mu_n = \frac{1}{|n|+ 1}$, then for each $n$ the vector $\sqrt{\mu_{n-1}}e_n\oplus \pm\sqrt{\mu_n}e_{n-1}$ is an eigenvector with eigenvalue $\pm\sqrt{\mu_n\mu_{n-1}}$.