Suppose the $f_i$ are dependent, so $\sum_i c_i f_i = 0$ for some scalars $c_i$. What does it mean for an element of $V^*$ to be zero? Well, such elements are defined to be linear functionals on $V$ and a function is zero if it takes the value zero everywhere on its domain. So for all $v\in V$ we have
\[
0 = \left(\sum_i c_i f_i\right)(v) = \sum_i c_i f_i(v) = \sum_i c_i\langle v,\alpha_i\rangle = \left\langle v,\sum_ic_i\alpha_i\right\rangle,
\]
where the first equation is the definition of a function being zero, the second is the definition of the vector space operations on $V^*$, the third is the definition of $f_i$, and the fourth is linearity of $\langle\cdot,\cdot\rangle$ in the second coordinate.
In particular, we can take $v = \sum_i c_i\alpha_i$, so we obtain $\left\langle \sum_i c_i\alpha_i,\sum_i c_i\alpha_i\right\rangle=0$. The positive definiteness property of an inner product says that the only $v$ with $\langle v,v\rangle=0$ is $v=0$, so $\sum_i c_i\alpha_i = 0$.
By assumption $\{\alpha_1,\ldots,\alpha_n\}$ is a basis for $V$, so it is linearly independent. Therefore $\sum_i c_i\alpha_i=0$ implies $c_i = 0$ for all $i$.
Since we started with the assumption that $\sum c_i f_i = 0$ and ended up with $c_i = 0$ for all $i$, we have shown that the set $\{f_1,\ldots,f_n\}$ is linearly independent as well. Since you know $\dim(V) = \dim(V^*) = n$, this means that $\{f_1,\ldots,f_n\}$ is a basis for $V^*$.
Yes, you can prove that via the bidual space $V^{**} = \mathrm{Hom}(V^*,K)$.
There is a canonical homomorphism $\iota: V \to V^{**}$ defined by $\iota(v)(f) := f(v)$ for $f \in V^{*}$. It is easy to see that $\iota$ is injective. Since $\dim V^{**} = \dim V^* = \dim V$, it is actually an isomorphism. Therefore, for every $\varphi \in V^{**}$ there exists a $v \in V$ s.t. $\varphi(f) = f(v)$ for $f \in V^*$.
Now, if $\{f_1,\ldots,f_n\}$ is a basis if $V^*$ then the elements $\{\varphi_1,\ldots,\varphi_n\}$ form a basis of $V^{**}$ where $\varphi_i(f_j) := \delta_{ij}$. By the above considerations, each $\varphi_i$ corresponds to an $\alpha_i \in V$ s.t. $f_i(\alpha_j) = \varphi_j(f_i) = \delta_{ij}$. Since $\iota: V\to V^{**}$ is an isomorphism and $\{\varphi_1,\ldots,\varphi_n\}$ is a basis of $V^{**}$ we know that $\{\alpha_1,\ldots,\alpha_n\}$ is a basis of $V$. Its dual basis is then $\{f_1,\ldots,f_n\}$ because $f_i(\alpha_j) = \delta_{ij}$.
Best Answer
Any two vector spaces with the same dimension are isomorphic as vector spaces, but there are many isomorphisms between them (choose a basis of one, and a basis of the other, and map the first basis to the second anyway you like). However, these isomorphisms all depend on a choice of basis.
So, if $V$ is finite dimensional, it is has the same dimension as $V^*$, so as vector spaces they are isomorphic.
As quid mentions in the comments, a finite-dimensional vector space $V$ is canonically isomorphic to its double-dual via $v\mapsto \hat{v}$ where $\hat{v}(f)=f(v)$ for all $f\in V^*$. Notice this is a vector space isomorphism independent of any choice of basis.