Example where $(V^*)^*\neq V$? Let $V$ be a vector space. This answer gives a nice explanation of why the "dual dual space" of $V$, i.e., $(V^*)^*$, is isomorphic to $V$ if $\dim V<\infty.$ 


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*Can someone give an example where $\dim V=\infty$ and $(V^*)^*\ncong V$? I'm having a hard time imagining this. 

*Additionally, must $V$ always be at least isomorphic to a subspace of $(V^*)^*$? It seems that the answer must be yes, since we can always define $\xi_{v∈ V}:V^*\to\mathbb R, \omega\mapsto\xi_v(\omega):=\omega(v).$
 A: Since you use the tag "functional analysis", I'll assume that you talk about continuous duals. 
Let $V=c_0$, that is the space of complex sequences that converge to zero. It is well-known (and fairly easy) to show that $V^*=\ell^1(\mathbb N)$, and $V^{**}=\ell^\infty(\mathbb N)$. 
In both cases the duality is $$\langle x,y\rangle=\sum_nx_ny_n.$$
The space $c_0$ is norm-separable, while $\ell^\infty(\mathbb N)$ isn't. 
As for your second question, yes, you always have a canonical injection $V\to V^{**}$. 

For completion's sake, the answer is similar if we consider the purely algebraic dual. Take $V=\mathbb R[x]$, the space of real polynomials. It has a countable basis, namely $\{x^n\}_{n\in\mathbb N\cup\{0\}}$. For each $t\in\mathbb R$ consider $\phi_t:\mathbb R[x]\to\mathbb R$ given by $\phi_t(p)=p(t)$. The uncountable set $\{\phi_t\}$ is linearly independent, so $\dim V^*>\dim V$. Then $\dim V^{**}\geq\dim V^*>\dim V$ and they cannot be isomorphic (the fact that the dimension of the dual is at least that of the space follows from considering dual basis). 
A: One can give a simple argument based on cardinality counting for the algebraic case.
Suppose that $V$ is a vector space of dimension $\mathrm{card}(I)$. Then $V$ is isomorphic to a direct sum $k^{\oplus{I}}$ of $I$-copies of $k$. Next
$$\left(k^{\oplus I}\right)^* = \mathrm{Hom}_k(k^{\oplus I},k) \cong k^{\prod I}$$
where the right hand side is the product of $I$ copies of $k$ (considered as a vector space over $k$). Suppose now that 
$$\mathrm{card}(k) < \mathrm{card}(I)$$
Then
$$\mathrm{card}\left(k^{\prod I}\right) \geq  \mathrm{card}\left(2^I\right) > \mathrm{card}(I) = \mathrm{card}\left(k^{\oplus I}\right)$$
This implies that under the assumption $\mathrm{card}(I) > \mathrm{card}(k)$ we have
$$\mathrm{card}(V^*) > \mathrm{card}(V)$$
In particular $\mathrm{dim}(V^*) > \mathrm{dim}(V) = \mathrm{card}(I)$. Now you can use exactly the same argument to show that
$$\mathrm{card}(V^{**}) > \mathrm{card}(V^*)$$
So in general $V$ and $V^{**}$ do not have the same cardinality.
A: For the first question. For every vector space $V$ it is true that 
$$
\dim(V)\leq \dim(V^*)
$$
with equality if and only if $V$ is finite dimensional (see Theorem 3.12 in Advanced Linear Algebra by Steven Roman). Thus, if $V$ is infinite dimensional, then $V\not\cong V^{**}$, so every infinite dimensional vector space serves you as an example.
For question 2: Yes.
