Canonical Isomorphism Between $\mathbf{V}$ and $(\mathbf{V}^*)^*$ For the finite-dimensional case, we have a canonical isomorphism between $\mathbf{V}$, a vector space with the usual addition and scalar multiplication, and $(\mathbf{V}^*)^*$, the "dual of the dual of $\mathbf{V}$." This canonical isomorphism means that the isomorphism is always the same, independent of additional choices. 
We can define a map $I : \mathbf{V} \to (\mathbf{V}^*)^*$ by $$x \mapsto I(x) \in (\mathbf{V}^*)^* \ \text{ where } \ I(x)(f) = f(x) \ \text{for any } \  f \in \mathbf{V}^*$$
My Question: what can go wrong in the infinite-dimensional case? The notes I am studying remark that if $\mathbf{V}$ is finite-dimensional, then $I$ is an isomorphism, but in the infinite-dimensional case we can go wrong? How?

There are two things that can go wrong in the infinite-dimensional (normed) case. 
First you could try to take the algebraic dual of $V$. Here it turns out that $V^{**}$ is much larger than $V$ for simple cardinality reasons as outlined in the Wikipedia article.
On the other hand, if $V$ is a normed linear space and you take the continuous dual, $V''$, then $V'$ (and thus also $V''$) will always be a Banach space. But! While $I$ as a map from $V$ to $V^{**}$ is obviously well-defined, it is not entirely obvious that $I(x)$ is in fact continuous for all $x$. In fact it is a consequence of the Hahn–Banach theorem which (roughly) states that there are "enough" continuous linear maps from $V$ to the base field in order for $V'$ and $V''$ to be interesting, e.g. the map $I$ is injective.
If $V$ is not a normed linear space, then things are more complicated and better left to a more advanced course in functional analysis.
