I am aware there are many proofs on this site of the fact that nilpotent operators are upper-triangularisable, but have been suggested an inductive proof that seems simpler than most others and wanted to check whether it is correct. The proof goes as follows:
Suppose we have $T:V\rightarrow V$ and $T^k = 0$ for some $k$. We will induct on the dimension $n$ of $V$.
$T$ has an eigenvalue zero, and so we can find an eigenvector $u$. Then $T\rvert_{V\setminus \langle u\rangle}$ also satisfies $(T\rvert_{V\setminus \langle u\rangle})^k=0$ and $V\setminus \langle u\rangle$ has dimension $n-1$, so by induction there is a basis $\mathcal{E}=\{e_2,\dots,e_n\}$ of $V\setminus \langle u\rangle$ such that $_\mathcal{E}[T\rvert_{V\setminus \langle u\rangle}]_{\mathcal{E}}$ is upper triangular.
With $u$, we get a basis $\mathcal{B} = \{u,e_2,\dots,e_n\}$ for $V$ such that
$$_{\mathcal{B}}[T]_{\mathcal{B}}=\begin{pmatrix}0 & * \\0 & _{\mathcal{E}}[T\rvert_{V\setminus \langle u\rangle}]_{\mathcal{E}} \end{pmatrix} \tag{*}$$
And so this matrix is upper triangular.
I feel like this proof doesn't make sense at all, given we don't even have that $V\setminus \langle u\rangle$ is a vector space, right?. I'm struggling to be able to explain this to them clearly though so any help would be appreciated.
The idea is nice, but there is a big hidden induction problem.
I translate what you tried to mean in more familiar words to me. Let me know if this was the same your idea.
If $T=0$, then it is clear. Hence assume $T\neq 0$. Let $K:=\ker (T)$ be the eigenspace of eigenvalue zero. Then we can complete $K$ to $V$ by considering a subspace $W$. Therefore, we have $V=K\oplus W$, and $W\neq 0$ since $T$ is different from zero.
Now we restrict $T$ on $W$. We would like to apply inductive hypothesis on $T_{|W}$, which has dimension less than $n$ since $K\neq 0$.
However, the induction hypothesis does not hold. Indeed, we firstly need to prove that $T_{|W}\colon W\to V$ is an endomorphism. The problem is that this never happens.
By contradiction, suppose that $T_{|W}\colon W\to W$ is an endomorphism. Then $T_{|W}^{k-1}\colon W\to W$ is an endomorphism. However, $T\circ T^{k-1}_{|W}=T^k=0$, therefore we have at the same time $T^{k-1}_{|W}(W)\subseteq W$ and $T^{k-1}_{|W}(W)\subseteq K$. Hence $T^{k-1}_{|W}(W)=0$ and so $T^{k-1}$ would be the zero map. This contradicts the fact that $k$ is the nilpotent ordering of $T$.
If you want, I can sketch you another proof, which does not use the induction argument :). Let me know.