proof that the dimension of a subspace is less than or equal to the dimension of the vector space

Question

Let $V$ be a finite-dimensional $K$-vector space and let $U$ be a subspace of $V$. I'm trying to prove that $\dim (U) \leq \dim (V)$, but I'm not sure if the arguments what I gave was enough. Please take a look at it: I supposed for the sake of contradiction that $\dim(U)>\dim(V)$, then I tried to write basis for both spaces $U$ and $V$ as $B_u=\{u_1, u_2, ..., u_m\}$ and $B_v=\{v_1, v_2, ..., v_n\}$ (respectively), where $m>n$. So, that means $B_u$ contains $m$ linearly independent vectors and $B_v$ contains $n$ linearly independent vectors. Since $U$ is a subspace of $V$ then $V$ contains $m$ linearly independent vectors. On the other hand, $B_v$ already contains all linearly independent vectors of $V$ which are $n$. Here is the contradiction.