Is it true that $\dim(X) \leq \dim(X^{\ast})$ for every infinite dimentional Banach space $X$? Given an arbitrary infinite dimensional Banach space $X$, can we deduce that it's dimension $\dim(X)$ (the cardinality of one of its Hamel bases) is less or equal of the dimension  $\dim(X^{\ast})$ of its dual space (the space of all continuous linear functionals $f:X\to\mathbb{R}$)?
 A: This is an interesting question which  has been unaddressed for a long time, so I'll give it a shot.
Lets denote the cardinality of a set $A$
by $|A|$. For a normed space $X$ we define its density
character $d(X)$ as the smallest cardinality of its dense sets, that
is $d(X)=\min\{|D|: D\subseteq X, \overline{D}=X\}$. In particular a separable normed space $X$ has $d(X)=\aleph_0$. 
We need the following three lemmas:

Lemma 1: If $X$ is an infinite dimensional vector space over $\mathbb{R}$ and
  $\dim X\geq |\mathbb{R}|$, then $\dim X=|X|$.

A proof of Lemma 1 can be found here. 

Lemma 2: 
  Let $X$, $Y$ be infinite dimensional Banach spaces with $d(X)\leq
d(Y)$. Then $|X|\leq |Y|.$

You can find a proof here.
For the last step, it is known from functional analysis that when
$X^*$ is separable, then so is $X$. If you check  the proof carefully you'll realise that    what is actually proven is the following: 

Lemma 3: Let $X$ be a normed space. Then $d(X)\leq d(X^*)$.
Proof.
  Let $\{x_a^*: a\in A\}$ be  a dense subset of $S_{X^*}$ of
  cardinality $|A|=d(X^*)$. For every $a\in A$ we pick an $x_a\in B_X$
  with $x_a^*(x_a)>\tfrac{1}{2}$ and set $Y= \langle \{x_a: a\in A\} \rangle$.
  The set $Y$ is dense in $X$: Otherwise, there would exist an $f \in
X^*$ with $\|f\|=1$ such that $f(y)=0$ for every $y\in Y$. But then
  for every $a\in A$,
  \begin{eqnarray*}
\|f-x_a^*\|\geq |f(x_a)-x_a^*(x_a)|=\frac{1}{2},
\end{eqnarray*}
  which implies that $f\notin \overline{\{x_a^*: a\in A\}}=S_{X^*}$, a
  contradiction. So $X$ contains a dense subset of cardinality
  $|Y|\leq |A|$, therefore $d(X)\leq |A|=d(X^*)$.

Combining the previous lemmas, we get an affirmative answer to remilt's question.
A: That is a great answer to a very interesting problem. Given your current solution, I would like to suggest a possible generalization to the case of normed spaces.
Since in the finite dimensional case everything is working well, let $X$ be an infinite dimensional normed space and consider the canonical embedding of $X$ to its double dual $X^{**}$ given by:
$T:X\rightarrow X^{**},\hspace{10pt}  T(x)=T_x \hspace{5pt}$ where for $x\in X,$ we define $T_x:X^*\rightarrow\mathbb{R}$ such that $ \hspace{7pt}  {T_x}(x^*)=x^*(x)$ 
Of course, T is an isometric embedding.
Consider the spaces:
$T(X)\hspace{5pt}$ (which is isometric to $X$ and obviously dense in $\overline{T(X)}$) and
$\overline{T(X)}\hspace{5pt}$ (which is a Banach space, as it is a closed subspace of the Banach space $X^{**}$)
(This process is of course standard when considering the completion of a normed space $X$).
Since $\overline{T(X)}$ is a Banach space, by your argument we must have that $\dim(\overline{T(X)})\leq \dim(({\overline{T(X)}})^*)$
But, since $T(X)$ is dense in $\overline{T(X)},$ we must have that the spaces ${(T(X))}^*$ and $({\overline{T(X)}})^*$ are isometric.
(Indeed, consider this result when stated in the following more general fashion:
Let $X$ be a normed space, $Z$ a dense subspace of $X$ and $Y$ a Banach space. Then, the spaces $\mathcal{B}(X,Y)$ and $\mathcal{B}(Z,Y)$ are isometric.
I will be glad to give hints to the proof of this fact, if anyone is interested).
Now combine all the previous arguments together to get:
$\dim(X)= \dim(T(X))\leq \dim({\overline{T(X)}})\leq \dim(({\overline{T(X)}})^*) = \dim(({T(X)})^*)= \dim(X^*)$
(Notice that we have also implicitly used the fact that if two spaces are isometric, then their duals must be isometric as well).
