Prove that for every $k$ there's an invariant subspace

Let $V$, a vector space above $\mathbb{C}$ and let $T:V\to V$, a linear transformation. Show that for every $0\le k \le n$ there is an invariant subspace of $T$ with a dimension $k$.

It seems like the right option proving it with an induction. The case of $k=0$ is trivial. Let's assume the claim is true for some $k > 0$. Hence, there's a $V_k$ such that $T(V_k) \subseteq V_k$ where $V_k$ is a subspace with the dimension $k$.

Hence, $V_k$ has a basis $B_k = v_1, \ldots, v_k$.

I don't know how to proceed. I thought maybe extending the basis of $V_k$ to a basis of $V$ but not sure if it would be helpful.

EDIT I may use the fact that since $\mathbb{C}$ is algebraically closed then there's a basis $B$ such that $[T]_B$ is an upper triangular matrix.

Since $\Bbb{C}$ is algebraically closed, you can always find eigenvalues of every endomorphism. You have to use induction on $n$.
For $n=1$, the thesis is obvious.
Now, suppose $n \ge 2$ and that you proved the thesis for $0, \dots, n-1$.
Let $W$ be an invariant subspace of dimension $1$ (this is spanned by any eigenvector $v_1$). Then $T$ induces an endomorphism $T'$ of $V/W$ defined as $$T'(v+W) = T(v)+W$$
Since $\dim (V/W) = n-1$, you can find a suitable basis of $V/W$, $\{ v_2+W, \dots, v_n+W \}$ by inductive hypothesis. Hence, you conclude by considering the good basis $v_1, \dots, v_n$.