If $W$ is a subspace of $V$, show that the basis for $W$ is a subset of the basis for $V$ Note: The dimensions here are not necessarily finite. The hint suggests to use Zorn's lemma, which states that if any chain in a family of sets has an upper bound, then that family of sets has a maximal element.
So I presume the idea here is to construct a chain using the bases of $W$ and $V$, and go about such a way to make the basis for $V$ a maximal element, thus the basis for $W$ is a subset for basis for $V$. However, when I attempt to put the pieces together, I seem to always fall into circular reasoning.
For example, let $w$ be the basis for $W$ and $v$ be the basis for $V$. Let $S_1 = w \cup v$. Let $S_2= w \cup v'$ where $v'$ is $v$ with the $i$th vector removed. Let $S_3 = w$. Then we have $S_3 \subset S_2 \subset S_1$. You can repeat the process of removing another vector from the union and construct a family of sets where each chain will have an upper bound of $S_1$. Therefore, by Zorn's lemma, this family of sets will have a maximal element. 
Now, I am not sure what new information I am given. How I interpret the result of Zorn's lemma is that $S_1$ is a maximal element. But this seems rather circular since I constructed it to be that way. I do not know what I can conclude from this method, or if it is even the correct approach.
Any help is appreciated, thank you.
 A: First of all, the title of your question needs to be slightly corrected:

If $W$ is a subspace of $V$, show that any basis for $W$ is a subset of a basis for $V$.

Since vector spaces have many bases, we can't speak of "the basis" unless we mean a specific one.
This is pretty important here. And it's related to one of the mistakes in your reasoning. The claim is that there exists some basis of $V$ that contains the given basis $w$ of $W$. But it's not any one. So starting with a certain basis $v$ of $V$ isn't going to work — simply because this $v$ may in fact not contain $w$ (some other basis does, but not this one), so you're trying to prove something that may not even be true. Instead, proving that such a $v$ exists should be the end result of this proof, not its starting point.
A pretty standard way to demonstrate that chains have upper bounds is by taking their unions. But before we get to that step, what family of sets are we talking about? Note that the family of sets doesn't have to be a chain itself. It can be however wild as a whole. For Zorn's Lemma, we just need to demonstrate that any chain picked from this family has an upper bound.
I can offer you step-by-step hints for a standard proof of this theorem.


*

*As our starting point, we have some basis $w$ of $W$.

*Let $\mathcal{S}$ be the family of all subsets of $V$ that are linearly independent and contain $w$, i.e. $\mathcal{S}=\{s\subseteq V \mid s\text{ is linearly independent}, w\subseteq s\}$.

*$\mathcal{S}$ is not an empty family, because $w\in\mathcal{S}$.

*Now demonstrate that if we take any chain of elements of $\mathcal{S}$, this chain has an upper bound in $\mathcal{S}$. Hint: this upper bound is the union of the sets in this chain.

*By Zorn's Lemma, $\mathcal{S}$ has a maximal element, call it $v$.

*Argue that this $v$ is what we want.

