How to prove that any Hamel basis of an infinite-dimensional separable real Hilbert space is uncountable? How to prove that any Hamel basis of an infinite-dimensional complete and separable (having a countable dense set ) real inner-product space is uncountable ? Do I have to use Baire-category theorem ? Please help 
 A: Since the space is infinite dimensional so its basis cant be finite.Now every complete metric space is of $2nd$ category and by Baire Category theorem it cant be expressed as a countable union of Nowhere dense sets .can you take it from here?
A: The open balls of radius $\frac{1}{2\sqrt2}$ around every point belonging the the orthonormal basis vectors are disjoint. A countable set can't intersect them all if there are uncountably many vector in you Hamel basis, so it is not dense, and then the space is not separable. So there is a problem with your set of hypothesis.
What preceeds is conditional to the fact that one can orthonormalize the Hamel basis, which we can :
Let me note $E$ your space. For a family $\{\,e_j\mid j\in J\,\}\subseteq E$ note $s(w)$ be the space of all $\sum_{j\in J}x_je_j$ with $\sum \lvert x_j e_j\rvert^2<\infty$ and with the set of $j$'s such $x_j\ne 0$ countable.
Call a "finite Gram-Schmidt" as a subset $J\subseteq I$ together with a total order $\le$ on $J$ and an orthonormal family $\{\,y_j\mid j\in J\,\}$ such that $y_j\in s(v)\cup\{\,y_k\mid k\in J, k<j\,\})$ for each $j\in J$.
The set of finite Gram-Schmidt is inductively ordered. Then Zorn's lemma ensures that this set a maximal element for the order. We must have $J=I$. (Do you see why ?)
We do have a Gram-Schmidt orthonormalization process in $E$. If we apply it to your Hamel base, we can suppose you Hamel basis to be orthogonal
