Dual Vector Space Proof I am trying to prove the following:

Let $\phi: V \rightarrow (V^*)^*$ be defined by $v \to (f \to f(v))$, where $v \in V$ and $f \in V^*$. Show that $\phi$ is an injective map which is an isomorphism if and only if $V$ is of finite dimension. Note that $V$ is a vector space. 

I am not too sure where to start on this at all, any help or hints would be appreciated.
Suppose for $v, v' \in V$ we have $\phi(v)=\phi(v')$, which means that these are equal as elements of $(V^\ast)^\ast$. So $$\forall f \in V^\ast: \phi(v)(f)= \phi(v')(f).$$

By definition of $\phi$, we know

$$\forall v \in V: \forall f \in V^\ast: \phi(v)(f)=f(v).$$

So $$\forall f \in V^\ast: f(v)= f(v').$$

If $v \neq v'$, find $f \in V^\ast$ such that $f(v) \neq f(v')$. 

It then follows that $v=v'$ and $\phi$ is injective.

If $\dim(V)=n$, then $V \simeq \mathbb{K}^n$ then $\dim(V)=\dim(V^\ast)$ so $V^\ast \simeq \mathbb{K}^n$. Repeat for $V^\ast$. This shows one direction.
Clarifications: 


I will replace your vector $v$ with $x$.
To the function $f\mapsto f(x)$ I will denote it by $\operatorname{ev}_x(f)=f(x)$.
Also, the dual space of $\textsf{V}^*$ I will denote it by $\textsf{V}^{**}$.


I hope everything that follows is helpful.

Lemma : Let $\textsf{V}$ a finite-dimensional vector space and let $x\in \textsf{V}$. If $\operatorname{ev}_x(f)=0$ for all $f\in \textsf{V}^*$, then $x=0$.

Or equivalently, if $x\neq 0$ then $\operatorname{ev}_x(f)\neq 0$ for some $f\in \textsf{V}^*$.

Proof : Let $x\neq 0$. Let $$\beta =\{v_1,v_2,\dots,v_n\}$$ be a basis for $\textsf V$ such that $v_1=x$. Now, let $$\beta^* =\{f_1,f_2,\dots,f_n\}$$ the dual basis of $\beta$. Then $f_1(v_1)=1\neq 0$. That is $$\operatorname{ev}_x(f_1)\neq 0$$ 
Finally, pick $f:=f_1$. $\blacksquare$

Now, with this lemma, suppose that $\textsf V$ is finite-dimensional and we will prove that $\operatorname{ev}_x$ is an injective map from $\textsf V$ to $\textsf{V}^{**}$.

Theorem : Let $\textsf V$ a finite-dimensional vector space and define the function $\phi : \textsf{V} \to \textsf{V}^{**}$ by $$\phi (x)=\operatorname{ev}_x$$
Then, $\phi$ is an isomorphism.

Proof : First, we see that $\phi$ is linear : 

Let $x,y\in \textsf V$ and $a$ an arbitrary scalar. For any $f\in \textsf{V}^*$ we have

$$\phi(x+y)(f)=f(x+y)=f(x)+f(y)=\phi(x)(f)+\phi(y)(f)=\left( \phi(x)+\phi(y)\right)(f)$$

and $$\phi(ax)(f)=f(ax)=af(x)=\left(a\phi(x)\right) (f)$$
So, $\phi$ is linear. Next, $\phi$ is one-to-one : 

Suppose $\phi(x)$ is the zero functional on $\textsf{V}^*$ for some $x\in \textsf V$ (it is the same as saying that $\phi(x)$ is the zero vector on $\textsf{V}^{**}$). Then $\phi(x)(f)=\operatorname{ev}_x(f)=0$ for all $f\in \textsf{V}^*$. By the previous lemma we conclude that $x=0$. 
That is $\phi$ is injective.

And since $\textsf V$ and $\textsf{V}^{**}$ has the same dimension, $\phi$ is also surjective. $\blacksquare$