How can I fix this proof using transfinite induction of the existence of bases of normed vector spaces? I want to prove that every normed vector space has a basis. The following proof relies on the principle of transfinite induction. I believe that it is flawed because I'm not so sure if it's possible to construct objects using transfinite induction like the ways it's possible using (finite) induction. Can you explain to me how it is flawed (if it actually is), and outline a way to fix it?
We will construct the collection $\{v_\alpha\}$, where $\alpha$ ranges over the ordinal numbers, of linearly independent subsets of $X$. Let $V_0 = \emptyset$, which is vacuously linearly independent.
Assume that $\{V_\beta\}_{\beta < \alpha_0}$ for fixed $\alpha_0 \ge 1$ has been so constructed. Form $V_{\alpha_0}$ as follows: 


*

*Choose any $v \notin \bigcup_{\beta < \alpha_0} \operatorname{span}(V_\beta)$ and let $V_{\alpha_0} = \{v\} \cup \bigcup_{\beta < \alpha_0} V_\beta$.

*If no such $v$ exists, then let $V_{\alpha_0} = \bigcup_{\beta < \alpha_0} V_\beta$.


Now let $\gamma$ be any ordinal of cardinality $|X|$. Then clearly $\operatorname{span}(V_\gamma) = X$, implying that $V_\gamma$ is a basis for $X$.
 A: First, as a side remark, note that 
$\bigcup_{\beta < \alpha_0} \operatorname{span}(V_\beta)= 
\operatorname{span}(\bigcup_{\beta < \alpha_0} V_\beta)$. 
The statement "Then clearly $\operatorname{span}(V_\gamma)=X$ " is flawed.
I'll explain why it is, and how to fix it.  
Say $B$ is a basis for $X$. One may easily come up with examples where $X$ is infinite and $|B|=|X|$ (e.g. take $X$ with large enough cardinality, larger that continuum, the latter being the cardinality of the reals). Say $\kappa=|X|=|B|$. It is known that $\kappa*\kappa=\kappa$ 
(this is analogous to the statement that the union of countably many countable sets 
is countable). In other words, it is possible to split $B$ into $\kappa$ many disjoint sets 
$B_\xi$, $\xi<\kappa$, each of cardinality $\kappa$. That is, $B$ is the disjoint union $B_0\cup B_1 \cup B_2 \cup \dots\cup B_\xi \cup \dots$, where 
$|B_0|=\kappa$, $|B_1|=\kappa$, and $|B_\xi|=\kappa$, for all $\xi<\kappa$. 
But the above means that it is conceivably possible that for the first $\kappa$ many steps in the construction that you describe (where now we interpret $\kappa$ as an ordinal), although we do add $\kappa$ many elements to our to-be-built basis, we only construct say $B_0$. Clearly $B_0$ on its own is not a basis, since the sets $B_1, B_2$, etc., were omitted. We have long way to go to make sure we get a basis for $X$. So, carrying on the construction for only $\kappa=|X|$ many steps need not necessarily produce a basis for $X$.  
How do things change is we keep going for $\kappa^+$ many steps (treating $\kappa^+$ as an ordinal) where $\kappa^+$ is the next cardinal after $\kappa=|X|$, that is, $\kappa^+$ is the smallest cardinal greater than $\kappa$ ? 
Well, it must be the case that the construction stabilizes before we go all the way to the top of $\kappa^+$. That is, there must be an ordinal $\mu<\kappa^+$ such that 
$X = \operatorname{span}(\bigcup_{\alpha < \mu} V_\alpha)$. For, if no such ordinal $\mu$ exists, then our construction would produce a 
$v=v_\alpha \notin \bigcup_{\beta < \alpha} \operatorname{span}(V_\beta)$ for every 
$\alpha<\kappa^+$. But since all these $v_\alpha$ are different, the set 
$\{v_\alpha:\alpha<\kappa^+\}$ would be a cardinality $\kappa^+$ set, contained in the set $X$ of cardinality $\kappa$, and this is a contradiction. So, there must be an ordinal $\mu<\kappa^+$ at which we have constructed our basis, and after which no new elements are added. Intuitively, since $X$ has only $\kappa$ many elements, it cannot supply enough material to build a set of cardinality $\kappa^+$, so the construction ought to run out of steam some time before it goes all the way up to $\kappa^+$.  
Just to avoid confusion, it is true that the cardinality of $\mu$ is $\kappa=|X|$. 
On other words, $|\mu|=\kappa$. But that does not mean that $\mu=\kappa$. It is possible that, 
as ordinals, we have $\kappa<\mu$. As a simplest illustration of this phenomenon (even if the cardinals involved are too small to be the cardinality of a basis) consider $\omega+\omega$ (ordinal addition) where $\omega$ is the smallest infinite ordinal. Here $\omega$ has the order type of the natural numbers, $0<1<2<...$, and 
$\omega+\omega$ has the order type $0<2<4<...<1<3<5<...$, i.e. all even natural numbers, followed by all odd natural numbers. As an ordinal, $\omega+\omega$ may be written as $0,1,2,...,\omega,\omega+1,\omega+2,...$. Clearly $\omega+\omega$ is a countable ordinal (just like $\omega$ itself), so $|\omega+\omega|=|\omega|=\aleph_0$ , yet, as ordinals, 
$\omega<\omega+\omega$, since $\omega=\{0,1,2,...\}$ is a proper initial segment of $\omega+\omega=\{0,1,2,...,\omega,\omega+1,\omega+2,...\}$. 
