Counterexamples for "every linear map on an infinite dimensional complex vector space has an eigenvalue" Every linear map on a finite dimensional complex vector space has an eigenvalue. Not so in the infinite case.
I'm interested in nice counterexamples anyone might have. 
Here's one:
Consider the vector space $\mathbb C^\infty$ of sequences and the right shift map $R$ defined by
$$R(a_1, a_2, a_3, ...) = (0, a_1, a_2, a_3, ...)$$
$R$ has no eigenvalue (using the usual convention that there must be a non-trivial eigenvector).

There is an easy generalisation of the example in the question. In the polynomial ring $\Bbb C[X]$, the $\Bbb C$-linear operator $\phi: Q\mapsto PQ$ of multiplication by a fixed non-constant polynomial $P$ cannot have any eigenvalues. This is simply so since if $Q\neq0$ were an eigenvector for$~\lambda$, then one would lave $(P-\lambda)Q=0$, contradicting that $\Bbb C[X]$, is an integral domain. The example in the question is just the case $P=X$.
You can generalise this further using other $\Bbb C$-algebras that are integral domains, of which there are many.
It might be worth noting, and I did in this answer, that though there are no eigenvalues, one can make (the image of) any nonzero vector become an eigenvector for any desired eigenvalue by taking a suitably chosen quotient of $\Bbb C[X]$ by a $\phi$-stable subspace.
