Basis is uncountable or countable when X is separable? The following is from Functional Analysis book by Conway:

Let $X$ be a separable infinite-dimensional Banach space and let ${\{e_i : i \in I}\}$ be a Hamel basis for $X$ with $|| e_i || = 1$ for all $i$. Note that a Baire Category argument shows that I is uncountable.

I am confused with two things:
1- How Baire Category Thm is related?
2- I think the book is wrong because I must be countable not uncountable? See this, for example.
 A: 
Let $X$ be a Banach space. Then, either $X$ has a finite Hamel basis (and in this case $X$ is finite-dimensional) or $X$ has an uncountable Hamel basis.

Proof:
First, note that using Zorn lemma, we can prove that every vector space has a Hamel basis. (If you need, I can add details here).
Let $X$ be a Banach space. To get a contradiction, let us suppose that $X$ has a countable infinite Hamel basis $B = \{ e_i \}_{i \in \Bbb N}$.
For each $n \in \Bbb N$, let $V_n$ be the subspace generated by $\{e_0, \cdots, e_n\}$. It is clear that $V_n$ is a finite dimensional subspace of $X$. So $V_n$ is closed.
Since, for all $n \in \Bbb N$, $V_n$ is a proper subspace of $X$, it is also clear that  $V_n$ has empty interior in $X$ (see Remark 2).
Now, because $B$ is a Hamel basis, we have that $X= \bigcup_{n \in \Bbb N} V_n$.
However, by Baire Category Theorem,  a complete metric space (in our case $X$) can not be the countable unions of closed subsets having empty interior (nowhere dense closed subsets). Contradiction. So $X$ can not have a countable infinite Hamel basis.
Remark 1:

*

*The fact that, for all $n \in \Bbb N$, $V_n$ is closed and has empty interior remains true even if $X$ is just an infinite-dimensional normed vector space.


*$X= \bigcup_{n \in \Bbb N} V_n$ is true because $B$ is a Hamel basis


*It is only to get the contradiction to Baire Category Theorem, that we need $X$ to be a Banach space (so $X$ is a complete metric space).
Remark 2: Let $X$ be any normed vector space and $V$ be any proper  subspace of $X$. Then $V$ has empty interior in $X$.
Proof: Let us prove the counter-positive. Suppose $V$ has an interior point $p$. It means that there is $\varepsilon >0$, such that $B_X(p, \varepsilon) \subseteq V$. Since $p \in V$ and $V$ is a vector space, we have that
$$  B_X(0, \varepsilon)= \{y - p : y \in B_X(p, \varepsilon) \} \subseteq V - p =V$$
Now, given any $x \in X$, let $K > \frac{ \|x\|}{\varepsilon}$. So
$ \|  K^{-1} x \| < \varepsilon $. So $ K^{-1} x  \in V$. Since $V$ is a vector space, $x = K(K^{-1} x) \in V$. So $X \subseteq V$. It means that $V$ is NOT a proper subspace of $X$.
So we have that, if $V$ is a proper  subspace of $X$, then $V$ has empty interior in $X$.
A: There is a dichotomy of the cardinality of Hamel bases. A Hamel basis of a Banach Space is either finite (then the Banach Space is finite dimensional) or it is uncountable (if the space is of infinite dimension) a prove of this using Baire's Category Theorem can be found here:
https://math.stackexchange.com/a/217843/860249
To get the notation straight: A Hamel basis is a set $E \subseteq X$ of a $\mathbb{K}$-Banach space $X$ which is linearly independent, i.e., for any distinct $x_1, \dots, x_n \in E$ and coefficients $c_1, \dots, c_n \in \mathbb{K}$ we have
$$ \sum_{k = 1}^n c_i x_i = 0 \implies c_1 = \cdots = c_n = 0, $$
and a generating set, i.e., for all $x \in X$ there exist $x_1, \dots, x_n \in E$ and coefficients $c_1, \dots, c_n \in \mathbb{K}$ such that
$$ x =  \sum_{k = 1}^n c_i x_i.$$
Notice that the sum here is always finite. With Schauder or Hilbert bases this need not be the case. Intuitively this requires that the Hamel bases to be larger, as less elements $x \in X$ can be represented using finite sums than using infinite series.
