What's wrong with this proof that all linear endomorphisms only have real eigenvalues? Let $f$ be an endomorphism in the real vector space $V$.
Let $\lambda=a+ib$ be an eigenvalue for $f$. We're going to show that $b=0$.
Let $v$ be an eigenvector relative to $a+ib$. Then
$f(v)=(a+ib)v=av+ibv$.
Since $f$ is in a real vector space, both $v$ and $f(v)$ must be real vectors, so all their components are real. In other words, $ibv=0$.
But $v \neq 0$, so $ib=0$ and $b=0$.

The only part that seems suspicious to me is when I say that $(a+ib)v=av+ibv$, because we know this distributive property holds for real scalars, being in a real vector space, and not necessarily for complex scalars - for all we know it doesn't even make sense to consider the object $(a+ib)v$.
 A: Example: Let $V= \mathbb R^2$ and let $f$ be given by the matrix $A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.
The eigenvalues of $A$ are $ \pm i$. There is no corresponding eigenvectors $v$ such that $v$ and $f(v)$ are elements of $ \mathbb R^2$ ......
A: If $V$ is a real vector space, the expression $\color{red}{(a+ib)\mathbf{v}}$ is simply undefined — there are no complex scalars in a real vector space. So from that point on, your "proof" is meaningless, unfortunately.
To make this meaningful, we can construct the complexification $V^{\mathbb{C}}$ of the real vector space $V$, see here. Then in $V^{\mathbb{C}}$, the expression $(a+ib)\mathbf{v}$ is fine, but it doesn't lead to any contradiction, as nothing has to be real anymore (so to speak).
Often, it seems pretty natural to involve complex eigenvalues even when $V$ is a real vector space. For example, if $V$ is a finite-dimensional vector space, then any linear transformation $f:V\to V$ can be represented with a square matrix, and its characteristic polynomial may have complex roots. But once these complex roots are introduced, we're effectively talking about the complexification $V^{\mathbb{C}}$, even if the author carelessly forgets to mention that.
