Does there exist a linearly independent and dense subset? Do there exist in infinitely dimensional normed spaces linearly independent and dense subsets?
(Existence of linearly independent dense subset is equivalent of existence of dense Hamel Basis.)
Thanks.
 A: I think I can sketch a proof that $\mathbb R$, viewed as a vector space over $\mathbb Q$, has a dense Hamel basis.  (Of course, this proof relies heavily on the Axiom of Choice; without it, $\mathbb R$ over $\mathbb Q$ might not have a Hamel basis at all.)
First, pick any Hamel basis $\mathcal H$ for $\mathbb R$ over $\mathbb Q$.  Let $f$ be an enumeration of $\mathbb Q$ (i.e. a bijection from $\mathbb N$ to $\mathbb Q$), and let $g$ be an injection from $\mathbb N$ to $\mathcal H$.
Now, for each $n \in \mathbb N$, choose an element $x_n$ from $g(n)\mathbb Q \cap [f(n)-\frac 1n, f(n)+\frac 1n]$.  The set $\mathcal X = \{x_1, x_2, \ldots\}$ is obviously linearly independent (over $\mathbb Q$). I claim it is also dense in $\mathbb R$.
To prove this, we need to show that every open interval $(a,b) \subset \mathbb R$, $a < b$, contains some $x_n \in \mathcal X$.  Let $\delta = (b-a)/3$, and let $m = 1/\delta$.  Since the interval $(a+\delta, b-\delta)$ contains infinitely many elements of $\mathbb Q$, it must contain some $q \in \mathbb Q$ such that $q = f(n)$ for some $n > m$.  Thus, $|x_n - q| \le \frac 1n < \frac 1m = \delta$, and so $x_n \in (a,b)$.
Of course, we can also extend $\mathcal X$ to a full Hamel basis by adding $\mathcal H \setminus g(\mathbb N)$ to it.
A: If your space (call it $X$) is separable then the answer is yes.
Pick a countable base $\{U_n\}$ for $X$.  Construct the sequence $\{x_1, x_2, \dots\}$ inductively as follows.  First choose $x_1 \in U_1$ arbitrarily.  Now for the inductive step, if $\{x_1, \dots, x_n\}$ are already chosen, set $E_n$ to be their linear span.  Since $E_n$ is finite dimensional, it is a proper subspace of $X$, and so it must have empty interior.  In particular it does not contain $U_{n+1}$, so choose $x_{n+1} \in U_{n+1} \backslash E_n$.
The sequence $\{x_1, x_2, \dots\}$ is linearly independent by construction, and dense because it intersects every $U_n$.
Note that we only used dependent choice, which is much less than is needed to produce a Hamel basis.  
The non-separable case seems more difficult.
Edit: After reading Martin Sleziak's answer I think I see how to handle the non-separable case, at the expense of more choice.
Fix a Hamel basis $\{e_i\}_{i \in I}$ for $X$, where $|I| = \kappa = \dim X$.  Let $A$ be the $\mathbb{Q}$-linear span of $\{e_i\}$; then $|A| = \kappa$ also.  $A$ is clearly dense in $X$, so $\mathcal{U} = \{B(x,r) : x \in A, r \in \mathbb{Q}\}$ is a base for $X$, and $|\mathcal{U}| = \kappa$ also.
Let $\omega$ be the smallest ordinal of cardinality $\kappa$.  We can then enumerate $\mathcal{U}$ as $\mathcal{U} = \{U_j\}_{j < \omega}$.  We now produce the desired set $\{x_j\}_{j < \omega}$ by transfinite induction: for $j < \omega$, let $E_j$ be the ($\mathbb{R}$-)linear span of $\{x_m : m < j\}$.  Then $\dim E_j = |j| < \kappa$, and in particular $E_j$ is a proper subspace of $X$.  As such it has empty interior, so does not contain $U_j$, and so we can pick $x_j \in U_j \backslash E_j$.  As before, the set $\{x_j\}_{j < \omega}$ is linearly independent by construction, and is dense because it intersects every $U_j$.
(Sorry if this is not written too clearly; I'm not really used to arguments like this.)
As in Martin's references, the key ingredient is a base of cardinality at most $\dim X$.  Of course there can be bases of smaller cardinality, as in the separable case, and you could use such a base to construct a smaller linearly independent dense set.
A: From the paper R. R. Phelps: Subreflexive normed linear spaces, Archiv der Mathematik, Volume 8, Number 6, 444-450

Theorem 3.1 (KLEE). Suppose the topological linear space $E$ has a neighborhood basis $\mathcal U$ at the origin such that $\operatorname{card} \mathcal U$ is less than or equal to the dimension of $E$. Then $E$ admits a dense Hamel basis.
A consequence of the above theorem is that every infinite dimensional metric
  linear space contains a dense Hamel basis. This result is given by MACKEY [8, p. 185] for $\aleph_0$-dimensional normed linear spaces.
G. W. Mackey, On infinite dimensional linear spaces. Trans. Amer. math. Soc. 57, 155--207 (1945).

A: This is quite an interesting question that I will attempt to answer even though I am not an expert in this field.
The first my thought was to say something about Fourier series or Stone-Weierstrass theorem, but then I realized the the question is whether such subsets exist in all normed spaces. 
After a second thought I recalled that these sets have a name Schauder bases. As it is well known, that Hamel bases exist in any vector space. The difference between these two types of bases is that a Hamel basis requires a finite linear combination to represent any element of the vector space, while with a Schauder basis one can take infinite sums provided there is a notion of convergence (as in normed spaces).
Then I did a little googling and I found out that the answer is NO.
According to this "Per Enflo found a separable Banach space that doesn't have a Schauder basis." (Per Enflo is a very interesting mathematician, by the way).
Later I came across another link where this topic was covered in a higher detail.
To @Davide Giraudo: A linearly independent subset is a subset that is not linearly dependent.
A subset of a vector space is called linearly dependent if there is an element in this subset which is a linear combination of a finite number of elements of this subset.
