How to show $(X^*)^*$ is finite dimensional when $X^*$ is of finite dimension? 
Let X be a normed $\mathbb K$-linear space .Is it true that if $X^*$ is finite dimensional then $(X^*)^*$ is also finite dimensional and then $X$ is finite dimensional?

Here $X^*$ is the dual space of $X$. 
Actually it is a part in a proof of a theorem. Here $X^*=span(\cup_{n\in \displaystyle{ \mathbb N} }f_n)$ and from here we conclude that $X^*$ is of finite dimensional as any Banach space is either finite or uncountable dimensional. Then the above question is written in the proof.
Please someone give some hints. to prove the above question.
Thank you. 
 A: If $X^*$  denotes the algebraic dual, or if $X^*$ is a locally convex space, it is an easy exercise that if $Y$ is finite-dimensional, then so is $Y^*$ (by constructing a dual basis). So in your case you know that $X^{**}$ is finite-dimensional. 
You also know that $X$ embeds as a subspace of $X^{**}$. As subspaces of finite-dimensional spaces are finite-dimensional, the conclusion is that $X$ is finite-dimensional. 
If $X^*$ denotes in general the continuum dual of a topological vector space, the result is not true. For example, it is well known that there exist infinite dimensional topological vector spaces with trivial dual.
A: Given a basis $e_1, \dots, e_n$ for the vector space $V$, define a set $\epsilon_1, \dots, \epsilon_n$ of elements of $V^*$ by $\epsilon_i(e_{ij}) := \delta_{ij}$. ($\delta_{ij}$ is equal to $1$ if $i=j$, and $0$ otherwise.)
Show that this set is independent, and that it spans $V^*$.
(Basically I've just taken the transpose of the fact that $e_1, \dots, e_n$ form a basis of $V$.)
