Let $V$ be a subspace of the complex vector space $M_n(\mathbb C)$.
Suppose that every non-zero element of $V$ is an invertible matrix.
Show that $\dim_{\mathbb C} V \leqslant 1$.
We know that, every $n\times n$ invertible is row equivalent to identity matrix this gives basis element of V has only one element i.e. identity matrix. hence the dimension of $V=1$. and if $\dim(V)=0$ the subspace is only zero matrices.
Please tell me if my argument or proof is correct or not.
No, it is not correct. Suppose, say, that$$V=\left\{\begin{bmatrix}z&0\\0&2z\end{bmatrix}\,\middle|\,z\in\mathbb C\right\}.$$Then every non-zero element of $V$ is invertible. But it is not true that the set $\{\text{identity matrix}\}$ is basis of $V$, since the identity matrix does not belong to $V$.