Below is a proof that any linear operator must have an eigenvalue. The proof obviously contains a mistake because the statement is wrong. But I do not see the mistake. Please point it out if you see it.
Assumption: $V$ is a finitely dimensional $K$-linear space ($K$ is a field with infinitely many elements, e.g. $\mathbb R$), $V \ne \{0\}$, $L:V \to V$ is a linear operator.
Claim: $L$ must have an eigenvalue $\lambda \in K$, i.e. there must be $\lambda \in K$ and $0 \ne v \in V$ such that $L(v)=\lambda v$.
Proof: Assume that $L$ does not have an eigenvalue. Then for all $\lambda \in K$ it should hold: $L-\lambda I$ is bijective ($I$ is the identity operator).
So we have an infinite family of bijective linear operators $S:=(L-\lambda I)_{\lambda \in K}$ in the linear space $H$ of linear operators $V\to V$ that is finitely dimensional ($\dim H = (\dim V)^2$).
$S$ is a linear independent family. Indeed, let $a_1,\ldots,a_n \in K$ such that $0=\sum_{i=1}^n a_i (L-\lambda_i I)=(\sum_{i=1}^n a_i)L - (\sum_{i=1}^n a_i\lambda_i) I$. If not all $a_i$ are $0$, the $S \ni L - \frac{ \sum_{i=1}^n a_i\lambda_i}{\sum_{i=1}^n a_i} I=0$. Therefore $S$ contains a zero operator. This contradicts the statement above that all elements in $S$ are bijective and $V \ne \{0\}$.
So we have found an infinite linearly independent family $S$ in a finitely dimensional linear space $H$. That is a contradiction. This $L$ has an eigenvalue.
Addition: $H$ is finitely dimensional
Proof: fix a basis of $V$ say $e_1,\ldots,e_m$. Consider linear operator $L_{i,j}$ for which hold $L(e_i)=e_j$ and $L(e_k)=0$ for $k \ne i$. Than $H$ is panned by the finite family $(L_{i,j})_{j,i = 1,\ldots m}$
The statement that $S$ is linearly independent is false. $\sum_{i=1}^na_i$ can be zero even if not all the $a_i$ are zero. As a specific example of how this fails, (over $\mathbb R$) set $\lambda_1=2,\lambda_2=3,\lambda_3=1$ with $a_1=1,a_2=-\frac 12,a_3=-\frac 12.$ Then $$a_1(L-\lambda_1 I)+a_2(L-\lambda_2 I)+a_3(L-\lambda_3 I)=0$$