I am trying to write a proof from the following statement:

Suppose that $V$ is a finite dimensional vector space over the field $\Bbb{F}$. Suppose that $W$ is a subspace of $V$. Show that $\text{dim}(W)\leq \text{dim}(V)$ and
$$\text{dim}(W)=\text{dim}(V)\iff W=V $$

My attempt is as follows:

Since $V$ is finite dimensional this implies that $$\exists\mathcal{B}\_V\,\text{such that}\,\text{dim}(V)=|\mathcal{B}\_V|=n,\,n\in\Bbb{N}$$

Since the basis of $V$ has a finite cardinality it is a finite spanning set of $V$. Because $\text{span}(\mathcal{B}\_V)=V$ and $W\leq V$ $\mathcal{B}\_V$ also spans $W$. All vector spaces have a basis so $\exists\mathcal{B}\_W$, where $|\mathcal{B}\_W|=\text{dim}(W)\leq|\mathcal{B}\_V|=\text{dim}(V)$.

This is as far is I've got. Any help completing this proof would be greatly appreciated.