Let do this by induction on the dimension:
- In dimension 1 it is true take $\{0\}\subset V$ with $\dim V=1$
- Let suppose the result is true for very subspaces $W$ of dimension $\dim U<n$ and let prove that is true for $V$ of dimension $\dim = n$ .
Since we are in complex space $T$ has eigenvalues and can decompose as direct sum of eigen-spaces that is
$$V= \bigoplus_{i =1}^{p} E_{\lambda_i}\equiv E_{\lambda_1}\oplus U$$
where we suppose that $T$ that has $p>1$ eigenvalues and let
$$U = \bigoplus_{i =2}^{p} E_{\lambda_i}$$
$$T =T_1\oplus T'$$
Where $T_1 =T|_{E_{\lambda_1}}$ and $T' =T|_{U}$ .
We asume $p>1:$
We recall and it is easy to show that $T( E_{\lambda_i})\subset E_{\lambda_i} $ since
$$T(\ker (T-\lambda_iI))\subset \ker (T-\lambda_iI)$$
Hence One see that $$ T( E_{\lambda_1})\subset E_{\lambda_1} $$
and
$$T( U)\subset U $$ since $p>1$
we have that $$r =\dim E_{\lambda_1} <n~~~and ~~~n-r =\dim U <n$$
Whence By asumption of induction,there are two chain
$$\color{green}{W_0⊆W_1⊆....⊆W_r=W}$$ such that $\dim W_i = i,~~i= 0,1,\cdots ,r$ and $T_1(W_i)\subset W_i$
and
$$\color{blue}{U_0⊆U_1⊆....⊆U_{n-r}=U}$$
such that $\dim U_i = i,~~i= 0,1,\cdots ,n-r$ and $T'(U_i)\subset U_i$
Now consider the chain
$$ \color{blue}{W_0\oplus U_0⊆W_0\oplus U_1⊆....⊆}\color{red}{W_0\oplus U_{n-r} =U\oplus W_0} \color{green}{\subset W_1\oplus U ⊆...⊆W_r\oplus U =W\oplus U=V}$$
That is $$ \begin{cases} \color{blue}{V_i~~~~~= W_0\oplus U_i}&\text{if}~~0\le i \le n-r\\
\color{green}{V_{n-r+i} = U\oplus W_i} &\text{if}~~0\le i \le r \end{cases} $$
- For $0\le i \le n-r$ we have
$\color{blue}{V_i =W_0\oplus U_i }$ then $\color{blue}{T(V_i) = T_1(W_0)\oplus T'(U_i) \subset W_0\oplus U_i = V_i}$
$$\color{blue}{\dim V_i = \dim U_i +\dim W_0 = i}$$
- For $0\le i \le r$ we have
$\color{green}{V_{n-r+i} =W_i\oplus U }$ then $\color{green}{T(V_i) = T_1(W_i)\oplus T'(U) \subset W_i\oplus U = V_{n-r+i}}$
and since,$\dim U=n-r$ and $\dim W_i =i$ we have
$$\color{green}{\dim V_{n-r+i} = \dim U +\dim W_{i} = n-r+i}$$
- We have the chain
$$V_0⊆V_1⊆....⊆V_{n}=V$$
- $T(V_i)\subset V_i$
The cases where $p= 1$ is obvious since in that cases, it means
$$V= \bigoplus_{i =1}^{p} E_{\lambda_i}=E_{\lambda_i} =\ker (T-\lambda_1 I)$$
i.e $$T= \lambda_1I$$
take $$V_i=\{v_1,v_2,\cdots,v_i\}$$
Where $\{v_1,v_2,\cdots,v_n\}$ is any basis of $V$