Recall that the dual space $V^{\ast}$ is the vector space of all linear transformations $T$ from $V$ to $\mathbb{R}$.
So what you're asking is given a $T \in V^{\ast}$, and $a \in \mathbb{R}$ does there always exist a vector $v \in V$ such that $T(v) = a$. This is equivalent to asking if there is a solution for every right hand side, or if your map $T$ is surjective. Now if $V$ is finite dimensional, we can apply the the Rank - Nullity Theorem:
If $T : V \longrightarrow \Bbb{R}$ for $V$ finite dimensional then $\dim V = \dim \ker T + \dim \operatorname{Im} T.$
Now if the dimension of the image is zero then this means that the dimensional of the kernel is equal to the dimension of $V$. Since $\ker T$ is a subspace of $V$, this means that $V = \ker T$ (exercise). In other words, $T$ is the zero map so we exclude this possibility.
Therefore this means that $1 \leq \dim \operatorname{Im} T \leq \dim \Bbb{R} = 1$ so that by the same reasoning as before and noting that the image of a linear transformation is always a subspace of the codomain that
$\operatorname{Im} T = \mathbb{R}.$
In other words, your map is surjective.