Showing that the following vectors are linearly independent in a subspace which they do not span. I am trying to better understand vector spaces and dimensions.
I could prove (i) via induction and the definition of linear independence?
However how can I approach the questions (ii),(iii) which consider finite dimensions?

 A: For (i), since you know that $\mathrm{span}\left\{\mathbb{w}_1,...,\mathbb{w}_n\right\}$ is a subspace of $W$ which does not span $W$, then there is some vector $\mathbb{w}_{n+1}\in W$ that is not in $\mathrm{span}\left\{\mathbb{w}_1,...,\mathbb{w}_n\right\}$. 
This means that we can not write $\mathbb{w}_{n+1}$ as a linear combination of the elements $\left\{\mathbb{w}_1,...,\mathbb{w}_n\right\}$, which means that $\mathbb{w}_{n+1}$ must be linearly indepenedent of $\left\{\mathbb{w}_1,...,\mathbb{w}_n\right\}$.
By the plus or minus theorem (you probably talked about a similar theorem in class), since  $\mathbb{w}_{n+1}$ is linearly independent of $\left\{\mathbb{w}_1,...,\mathbb{w}_n\right\}$ we can add it to the set, resulting in $\left\{\mathbb{w}_1,...,\mathbb{w}_n,\mathbb{w}_{n+1}\right\}$ which will also be a linearly independent set.
For (ii), let $\dim V=n$. Clearly if $W=V$, then $\dim V=\dim W$, which is trivial. If $W$ is not a trivial subspace, then we know that $W$ is a closed subspace that does not span $V$. From there we can deduce that $\exists \mathbb{v}\in V$ that is not in $\mathrm{span}\ W$.
We can think of $\dim V$ as the number of elements in a basis for $V$. Similarly, we can think of $\dim W$ as the number of elements in a basis for the subspace $W$. Since $W$ does not span $V$, and we claimed that $\exists \mathbb{v}\in V$ that is not in $\mathrm{span}\ W$, then we can not write $\mathbb{v}$ as a linear combination of basis elements of $W$. From this, we can reason that we must add at least one vector to a basis of $W$ to get a basis for $V$. Thus, for any nontrivial subspace $W$ of $V$, we must have $\dim W < \dim V$.
Thus, $\dim W \leq \dim V$.
For (iii): We have that $W_1\subseteq W_2$ and $\dim W_1 = \dim W_2=n$. We know that $W_1$ has $n$ elements in a basis. Similarly, we know that $W_2$ has $n$ elements in a basis. Since $W_1\subseteq W_2$, then every element spanned by the basis elements of $W_1$ must be in $W_2$, thus, we can use the basis elements of $W_1$ to construct a basis for $W_2$. But we know that $\dim W_2=n$, thus the basis for $W_1$ containing $n$ elements must also be a basis for $W_2$. If we call this basis $B$, then we have $W_1=\mathrm{span}\ B=W_2$. So the result holds.
