Every $k$-dimensional subspace is $T$-invariant $\implies T= \lambda I$

Assume $\dim(V) = n$, and if possible, let there exist a $v\in V$ such that $Tv$ is linearly independent of $v$. Then $\{v, Tv\}$ is an independent set, then we can extend this basis to a basis of $V$. Let $S =\{v,Tv,v_1, \cdots, v_{n-2}\}$ is a basis of $V$.

Now let $W = span\{v,v_1, \cdots, v_{k-1}\}$, which is a subspace of dimension $k$ for each $k \in \{1,2, \cdots, n-1\}$, which contains $v$ but not $Tv$. Which is a contradiction, as each subspace of dimension $k$ has to be $T$ invariant. So $v$ is dependent on $Tv \implies Tv = \lambda_v v$. Now, as we take this $v \in V$ is arbitrary, so this is true for all $v \in \mathbb{F}$.

Now let $v,w$ be two independent vectors, then $T(v+w) = \lambda_{v+w}(v+w) = \lambda_vv + \lambda_ww \iff \lambda_v = \lambda_{v+w} = \lambda_{w}$. So there must be a unique scalar for all vectors in $V$.

Which finally shows $T = \lambda I$.

Based on the wording, indeed $k$ is fixed.

Hint 1: Show that if it holds for some $k\geq 1$, then it holds for $k=1$.

If $U_1,U_2\subset V$ are $k$-dimensional subspaces with $T(U_i)\subseteq U_i$ then what can you say about $T(U_1\cap U_2)$?

Hint 2: Show that all $1$-dimensional subspaces are scaled by the same factor.

If $u_1,u_2\in V$ and $T(u_i)=\lambda_i u_i$, then what can you say about $T(u_1+u_2)$?