Let $G$ be a closed subgroup of $\textrm{GL}_n$, let $V = k^n$. I'm trying to show:
(i): If $W$ is a subspace of $V$ stable under the action of $G$, then $W$ is also stable under the action of the Lie algebra $\mathfrak g$, where we interpret $\mathfrak g$ as a Lie subalgebra of $M_n$.
(ii): If $v \in V$, and $xv = v$ for all $x \in G$, then $Xv = 0$ for all $X \in \mathfrak g$.
For (i), we can extend a basis for $W$ to a basis for $V$. Changing bases (that is, conjugating by some $y \in \textrm{GL}_n$), we can assume $W = k^t \times \{0\} \times \cdots \times \{0\}$ for $1 \leq t < n$. In that case, $T_{ij} : t+1 \leq i \leq n, 1 \leq j \leq t$ are contained among a set of generators for the vanishing ideal $I$ of $G$ in $k[\textrm{GL}_n]$.
Now the Lie algebra $\mathfrak g$ of $G$ consists of all $X \in M_n$ which send $I$ to zero. Here $X$ can be interpreted as a function on $k[\textrm{GL}_n]$ by $X(T_{ij}) = X_{ij}$. It follows that all the $X \in \mathfrak g$ satisfy $X_{ij} = 0$ for $t+1 \leq i \leq n , 1 \leq j \leq t$, which gives us (i).
I have tinkered around with (ii) for awhile, but I haven't gotten anywhere. I hope someone might be able to think of a good hint. We know from (i) that $Xv = \lambda v$ for some $\lambda \in k$, but we need to show $\lambda = 0$.
Well now, that I think about it, I can do the same kind of trick I just did. Reduce to the case where $$v = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0\end{pmatrix}$$ And the fact that $xv = v$ for all $x \in G$ implies that $x_{11} = 1$ for all $x \in G$. So $T_{11} -1$ is contained in a set of generators for the vanishing ideal of $G$. If $X \in \mathfrak g$, then $0 = X(T_{11} -1) = X_{11} - X(1) = X_{11} - 0$, because derivations take the value $0$ on constants.
So $\mathfrak g$ consists of matrices with $0$ in the top left entry, done.