Exercise: Prove or give a counterexample: if $V$ is finite-dimensional and $U$ is a
subspace of $V$ that is invariant under every operator on $V$, then $U = \{0\}$
or $U = V$.
Operator: The term operator here is used to mean a linear map from a vector space to itself.
Proof: We will prove the contrapositive. Suppose that $U\ne \{0\}$ and $U\ne V$. Because every subspace of a finite dimensional vector space $V$ is part of a direct sum equal to $V$, there exists a subspace $W$ of $V$ such that $V=U\oplus W$. Let $u_1,\dots,u_m,w_1,\dots,w_n$ be a basis of $V$ consisting of the basis of $U$ and the basis of $W$. Define $T : V\to V$ by $$Tu_i=u_i \text{ for } i=1,\dots,m-1$$ $$Tu_m=w_1$$ $$Tw_i=w_{i+1} \text{ for } i=1,\dots,n-1$$ $$Tw_n=w_n$$.
Clearly $T$ is an operator on $V$ but $T$ is not an operator on $U$ as $T|_U(u_m)\notin U$.
Is the proof correct?
Edit: Upon the suggestions in the comments, I thought of typing up the proof in a more correct way for any future readers. Using the same ideas but stating where the hypothesis is being used and defining $T$ in a simple way.
Proof $2$: Again we will be proving the contrapositive. Suppose that $U\ne \{0\}$ and $U\ne V$. Because every subspace of a finite dimensional vector space is part of a direct sum equal to it, there exists a subspace $W$ of $V$ such that $V=U\oplus W$. Let $u_1,\dots,u_m,w_1,\dots,w_n$ be a basis of $V$ consisting of the basis of $U$ and the basis of $W$. The $u's$ in the basis of $U$ exist as $U\ne \{0\}$ and this basis can be extended to a basis of $V$ by adding additional vectors because $U\ne V$.
Define a linear map $T : V\to V$ by $$T(a_1u_1+\dots+a_nw_n)=a_nw_n$$ Then we have that $T(u_1)\notin U$. Hence, $T$ is not an operator on $U$.
Best Answer
Your proof is correct but requires reader to know about direct sum notion and this result. Proof: Assume towards contradiction, $\exists$ $U$ subspace of $V$ such that $U\neq \{0\},V$ and $U$ is invariant under $T$, $\forall T\in L(V,V)$. Let $\{\alpha_1,…,\alpha_r\}$ be basis of $U$, where $1\leq r\lt n$. Let $\{\alpha_1,…,\alpha_n\}$ be basis of $V$. Define linear operator $T$ on $V$ such that $T(\alpha_i)=\alpha_n$, for all $\leq i\leq n$. Then $T(\alpha_1)=\alpha_n\notin \text{span}(\{\alpha_1,…,\alpha_r\})=U$. So $U$ is not invariant under $T$. Thus we reach contradiction. Hence $U=\{0\}$ or $U=V$.
There are probably thousand way to construct linear operator to reach contradiction.