= Solution
The <dual space> here is the algebraic <dual space> $X^*=\operatorname{Hom}_{\mathbb R}(X,\mathbb R)$, consisting of every <linear functional>; no <topology> or boundedness is imposed. For a <basis> $e_1,\ldots,e_n$, define $e_i^*(\sum_jx_je_j)=x_i$. These are <linear functionals> satisfying $e_i^*(e_j)=\delta_{ij}$. Evaluating $\sum_i a_ie_i^*=0$ on each $e_j$ gives $a_j=0$, proving <linear independence>. Conversely, for every $T\in X^*$,
$$
T=\sum_{i=1}^nT(e_i)e_i^*.
$$
This proves spanning and hence proves, without assuming a dimension theorem, that the <dual basis> is a <basis> and $\dim X^*=n$.
The canonical <bidual evaluation map> is
$$
\boxed{J:X\longrightarrow X^{**},\qquad J(x)(T)=T(x)}.
$$
Its definition makes no choice of <basis> and it is <linear>. If $x\ne0$, one coordinate functional is nonzero on $x$, so $J$ is <injective>. For any $B\in X^{**}$, choose $x=\sum_iB(e_i^*)e_i$. Expanding $T$ in the <dual basis> gives $T(x)=\sum_iT(e_i)B(e_i^*)=B(T)$, proving <surjection> explicitly. Thus $J$ is an <isomorphism>.
If $T_1,\ldots,T_n$ is any <basis> of $X^*$, its <dual basis> $B_1,\ldots,B_n$ in $X^{**}$ exists by the result just proved. Put $v_j=J^{-1}(B_j)$. These vectors form a <basis> of $X$, and $T_i(v_j)=B_j(T_i)=\delta_{ij}$. \b[Every <basis> of the finite-dimensional dual is a dual <basis>], of a uniquely determined <basis> of $X$.
For a <separating subspace of an algebraic dual> $W$, choose a <basis> $T_1,\ldots,T_r$ of $W$. The <linear map> $x\mapsto(T_1(x),\ldots,T_r(x))$ is <injective> because $W$ separates nonzero vectors. The <rank-nullity theorem> implies $n\leq r$; but $W\subseteq X^*$ implies $r\leq n$. Hence
$$
\boxed{W=X^*}.
$$
The zero-dimensional case also has this conclusion.
For $X=\mathbb R[x]$, the <monomials> $1,x,x^2,\ldots$ form an infinite <basis>, each <polynomial> using only finitely many of them. The map $T\mapsto(T(1),T(x),T(x^2),\ldots)$ identifies $X^*$ with all real <sequences>, because any sequence $(a_j)$ defines $T(\sum_{j=0}^d c_jx^j)=\sum_{j=0}^dc_ja_j$. The <linear subspace> of finitely supported sequences is proper (it excludes $(1,1,\ldots)$) and separating (one coefficient functional detects any nonzero <polynomial>). Thus
$$
\boxed{\mathbb R^{(\mathbb N_0)}\subsetneq\mathbb R^{\mathbb N_0}\cong X^*\text{ is separating}}.
$$
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