Suppose $V$ is finite-dimensional and $E$ is a subspace of $\mathscr L(V)$

You can choose the basis you like. Let $T\ne0$, $T\in E$. Suppose $T(v_1)\ne0$, so in particular $v_1\ne0$.

Complete $v_1$ to a basis $\{v_1,\dots,v_n\}$.

Since $w_1=T(v_1)\ne0$, you can complete it to a basis $\{w_1,\dots,w_n\}$. Consider the maps $R_i$ and $S_i$ defined by $$ R_i(v_j)=\begin{cases} v_1 & j=i\\[4px] 0 & j\ne i \end{cases} \qquad S_i(w_j)=\begin{cases} v_i & j=1 \\[4px] 0 & j>1 \end{cases} $$ Then $$ S_i T R_i (v_i)=S_i T(v_1)=S_i(w_1)=v_i $$ and, for $j\ne i$, $$ S_i T R_i (v_j)=S_iT(0)=0 $$ Hence, if $U=\sum_{i=1}^n S_iTR_i\in E$, we have $$ U(v_j)=v_j $$ for $j=1,\dots,n$. So $U$ is the identity and therefore $E=\mathscr{L}(V)$.