[proofplan]
Choose an ordered basis of $V$ and represent $T$ by a complex square matrix $A$. The characteristic polynomial $p(\mu)=\det(A-\mu I_n)$ has positive degree because $V$ is nonzero, so the Fundamental Theorem of Algebra gives a complex root $\lambda$. At that root the matrix $A-\lambda I_n$ is singular, hence the corresponding operator $T-\lambda \operatorname{id}_V$ has a nonzero kernel vector, which is exactly an eigenvector.
[/proofplan]
[step:Represent the operator by a complex matrix in a chosen basis]
Let $n := \dim_{\mathbb{C}} V$. Since $V$ is finite-dimensional and nonzero, $n \in \mathbb{N}$. Choose an ordered basis $\mathcal{B} = (e_1,\ldots,e_n)$ of $V$ over $\mathbb{C}$.
Let $A \in \mathbb{C}^{n \times n}$ be the matrix of $T$ with respect to $\mathcal{B}$, meaning that for every $v \in V$,
\begin{align*}
[T(v)]_{\mathcal{B}} = A [v]_{\mathcal{B}}.
\end{align*}
Let $I_n \in \mathbb{C}^{n \times n}$ denote the identity matrix, and let $\operatorname{id}_V: V \to V$ denote the identity map on $V$.
[/step]
[step:Find a complex root of the characteristic polynomial]
Define the characteristic polynomial associated to $A$ as the map
\begin{align*}
p: \mathbb{C} \to \mathbb{C}, \qquad \mu \mapsto \det(A-\mu I_n).
\end{align*}
As a polynomial in $\mu$, the term obtained by selecting $-\mu$ from every diagonal entry in the determinant expansion is $(-1)^n \mu^n$, and every other term has degree at most $n-1$. Hence $p$ has degree $n \geq 1$.
By the Fundamental Theorem of Algebra, the nonconstant complex polynomial $p$ has a root. Therefore there exists $\lambda \in \mathbb{C}$ such that
\begin{align*}
p(\lambda)=0.
\end{align*}
[guided]
The nonzero hypothesis on $V$ is used here. Since $V$ is nonzero and finite-dimensional, its dimension $n := \dim_{\mathbb{C}} V$ is a positive integer. After choosing an ordered basis $\mathcal{B}$, the operator $T$ is represented by a matrix $A \in \mathbb{C}^{n \times n}$.
We define
\begin{align*}
p: \mathbb{C} \to \mathbb{C}, \qquad \mu \mapsto \det(A-\mu I_n).
\end{align*}
This is a complex polynomial. Its leading term is found from the determinant expansion: the product of the diagonal terms $-\mu$ contributes $(-1)^n\mu^n$, while any determinant term using at least one entry of $A$ contains at most $n-1$ factors involving $\mu$. Thus $p$ has degree $n$, and since $n \geq 1$, it is nonconstant.
The Fundamental Theorem of Algebra applies to every nonconstant polynomial with complex coefficients. Therefore there exists $\lambda \in \mathbb{C}$ such that
\begin{align*}
p(\lambda)=0.
\end{align*}
This root is the candidate eigenvalue.
[/guided]
[/step]
[step:Convert the root into a nonzero kernel vector]
From $p(\lambda)=0$, we have
\begin{align*}
\det(A-\lambda I_n)=0.
\end{align*}
Therefore the matrix $A-\lambda I_n$ is singular. Hence the linear map on $\mathbb{C}^n$ represented by $A-\lambda I_n$ has nonzero kernel, so there exists $x \in \mathbb{C}^n$ with $x \neq 0$ and
\begin{align*}
(A-\lambda I_n)x = 0.
\end{align*}
Since $\mathcal{B}$ is a basis, there is a unique $v \in V$ such that $[v]_{\mathcal{B}}=x$. Because $x \neq 0$, this vector satisfies $v \neq 0$. Using the definition of $A$ and the coordinate representation of the identity map, we obtain
\begin{align*}
[T(v)-\lambda v]_{\mathcal{B}} = (A-\lambda I_n)[v]_{\mathcal{B}}.
\end{align*}
Substituting $[v]_{\mathcal{B}}=x$ gives
\begin{align*}
[T(v)-\lambda v]_{\mathcal{B}} = (A-\lambda I_n)x = 0.
\end{align*}
Since coordinate representation in a basis is injective, $T(v)-\lambda v=0$, and therefore
\begin{align*}
T(v)=\lambda v.
\end{align*}
Thus $\lambda$ is an eigenvalue of $T$ with eigenvector $v$.
[/step]
