Theorem:
Let $V$ be an vector space. Then the dual space of $V$'s dual space is canonically isomorphic to $V$.
I am able to prove that $V$ is a subspace of $V^{**}$, the map $\mathbf{v}(\mathbf{v}^*) = \mathbf{v}^*(\mathbf{v}), \mathbf{v} \in V, \mathbf{v}^* \in V^*$ is injective, and $\mathrm{dim}(V) = \mathrm{dim}(V^*) = \mathrm{dim}(V^{**})$. I don't know how to continue from this point.
This only follows when $V$ is finite dimensional. The proof then follows from the rank-nullity theorem. So we have $T:V\to V^{**}$ by $T(v)=ev_v$, the evaluation map, i.e. $ev_v(f)=f(v)$ for $f\in V^*$. Then $T$ is injective: Suppose $T(v)=ev_0=ev_v=0$, or evaluation at $0$. If $v\not=0$, then we can find an $f\in V^*$ such that $f(v)\not=0$ (complete $\{v\}$ to a basis $\{v,v_1,\dots, v_{n-1}\}$, then define $v^*\in V^*$ by $v^*(v)=1$, $v^*(v_j)=0$ for all $j$). Then $ev_v(f)=f(v)\not=0=f(0)=ev_0$. Thus $T(v)=0\iff v=0$ so $T$ is injective. Then by Rank-Nullity
$$\dim V=\dim \text{Im}(T)+\dim\text{ker}(T)=\dim V^*=\dim V^{**}$$
By the fact that, for finite dimensional vector spaces, $\dim V=\dim V^{*}$.
There is a subtlety here: the isomorphism $V\to V^{*}$ and $V^{\ast}\to V^{\ast\ast}$ are not canonical (depend on basis choice), as does the proof I provided that for $v\in V, v\not=0$, there is $f\in V^*$ such that $f(v)\not=0$. However, combining everything doesn't require the choice of a basis, just that one exists.
An counterexample is provided here in the case that $V$ is not finite dimensional.
Let $l_0$ be the set of real sequences $(r_n)_{n \in N}$ such that $\lim_{n \to \infty}r_n=0$ with norm $||(a_n)_{n \in N} ||= \sup \{ |a_n \}_{n \in N}$.Then $l_0^*=l_1$ is the set of real sequences $q= (q_n)_{n \in N}$ satisfying $||q||= \sum_{n \in N} |a_n|< \infty $... [ For $ r=(r_n)_n \in l_0$ and $q=(q_n)_n \in l_1$,we have $q(r)=\sum q_np_n ]$... And $l_0^{**}=l_1^*=l_\infty $ is the set of all bounded real sequences $(s_n)_{n \in N}$ with $||(s_n)_n||= \sup_n |s_n|$. The canonical embedding of $l_0$ into $l_0^{**}$ is obviously not a surjection, so it's not an isomorphism.
