I want to prove this classic theorem of linear algebra:
Let $V$ be a finite dimensional vector space, with $\mathrm{dim}(V)=n$. Let $U$ be a subspace of $V$ with $\mathrm{dim}(U)=m$, then $m\leq n$ and if $m=n$ then $U=V$.
I know there is the standard proof (e.g. Friedberg), who builds a maximal set of linearly independent vectors of $U$, then arguing that none linearly independent set of $V$ can contain more than $n$ elements, he concludes that this maximal set is a basis of $U$ and that it contains $m\leq n$ elements.
My alternative attempt: I was thinking about a simpler proof, so I thought in something like this.
Suppose $m\geq n$. Then a basis of $U$ has $m$ elements and a basis of $V$ has $n$ elements. Then there exists at least one vector $x \in U$ but $x \notin V$. Now, by the definition of subspace, $U$ must be a subset of $V$. Which is a contradiction due to the existence of at least this one element $x$. So $m\leq n$.
Question Is my attempt valid or at least the idea is correct? How can I improve it?