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To find the basis of a dual space, one needs to solve equations to satisfy Kronecker's delta conditions. One we solve them we have everything needed to prove their uniqueness, linear independence, etc..

So my question is why those n² linear equations are solvable in the first place?

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    What equations are you talking about? The linear functionals $\;\phi_{i}\;$ are *defined* such that $\;\phi_i v_j=\delta_{ij}\;$ **once a basis** $\;\{v_j\}\;$ is already given in the vector space... – 2017-01-06
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    I am talking about those equations that define ϕ-s, that you have written. Why should they be solvable? Maybe just because v-s are basis vectors and are linearly independent? Is that enough to claim they are solvable? I am not sure – 2017-01-06

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Given a basis for a vector space, we can define linear functions on the vector space simply by picking any image for each basis element.

That's all these equations are: specifying the values of linear functionals on the elements of the basis.

We could instead give a pointwise definition of these functionals

$$ \phi_i\left( \sum_j c_{ij} v_j \right) = c_{ii} $$

but that's more complicated and has a lot of unnecessary redundancy.

In coordinates, it's just as trivial; the coordinate form of the basis vector $v_j$ is just the standard $n \times 1$ basis column vector $\hat{e}_j$. The coordinate form of covectors in the dual space are $1 \times n$ row vectors, with $\phi_i$ corresponding to the standard basis row vector $\hat{e}_i^T$.

The equations are just computing the products $\hat{e}_i^T \hat{e}_j$, and give a coordinate-free way to express the relationship between the two bases.