$V$ is isomorphic to $V^{\ast\ast}$, the double dual space of $V$.

Denote the evaluation map at $v$ by $\bar{v}$. Then the map $\tau v=\bar{v}$ from $V\to V^{\ast\ast}$ is indeed an injective linear transformation. Linearity is easy. Now suppose $v\in\ker(\tau)$. Then $$ \begin{align*} \tau v=0 &\implies \bar{v}=0\\ &\implies \bar{v}(f)=0\quad\forall f\in V^\ast\\ &\implies f(v)=0\quad\forall f\in V^\ast\\ &\implies v=0 \end{align*} $$ since $v\in V$ is zero iff $f(v)=0$ for all linear functionals on $V$. Thus $\ker(\tau)=\{0\}$, so $\tau$ is injective.

To see it is an isomorphism, it is useful to recall the fact that $$ \dim(V)\leq\dim(V^\ast) $$ with equality holding iff $V$ is finite dimensional, so applying this twice you find $$ \dim(V)=\dim(V^\ast)=\dim(V^{\ast\ast}) $$ so $\tau$ is an injective transformation between vector spaces of equal dimension, hence and isomorphism by rank-nullity.

Without knowing where you are stuck, it is difficult to give informative answers. I'll give two hints:

• If a vector space homomorphism between vector spaces of the same finite dimension is injective, then it is also $---$?
• To show a map between vector spaces in injective, show that if $v$ maps to zero, it implies $v$ zero too.