Discrete Fourier Transform: Prove that only N distinct values of {X(k)} can be computed
It's not quite clear what you are supposed to consider as starting point for your proof. You have to make some assumptions of what is known before you can prove something else. So unless you give a more detailed description of what you want to prove from what, I will give you a proof that uses a sensible set of assumptions.

Let's characterise the discrete Fourier transform as a map between two complex vector spaces $$\mathcal{F}:\mathbb{C}^N\to\mathbb{C}^M$$ of dimension $N$ and $M$ respectively. The map is supposed to preserve the structure of the vector space, so it must be linear.

A linear map between finite dimensional vector spaces can be completely described using a matrix, which in this case is $M\times N$. The rank of this matrix $F_+$ is less or equal to the smaller dimension, i.e. $\mathrm{rank}(F_+)\le\min(N,M)$.

We also require the map to be a transform, meaning that it can be inverted. So $$\mathcal{F}^{-1}:\mathbb{C}^M\to\mathbb{C}^N$$ must exist and $$\mathcal{F}^{-1}\circ\mathcal{F}=\mathrm{id}_N$$ as well as $$\mathcal{F}\circ\mathcal{F}^{-1}=\mathrm{id}_M$$
and the matrix $F_-$ of the inverse transform is also less or equal to the smaller dimension, i.e. $\mathrm{rank}(F_-)\le\min(N,M)$.

We can write the inversion conditions as a matrix equation
$$F^{-1}\cdot F = E_N$$
$$F\cdot F^{-1} = E_M$$
where $E$ is the identity matrix of the size specified in the subscript. The matrix multiplication can be translated to a rank inequality:

$$\min\left(\mathrm{rank}(F_-),\mathrm{rank}(F_+)\right)\geq\mathrm{rank}(E_N)=N$$
and
$$\min\left(\mathrm{rank}(F_+),\mathrm{rank}(F_-)\right)\geq\mathrm{rank}(E_M)=M$$
because the product of two matrices has at most the rank of the matrix with the lesser rank.
From above we also had 
$$\mathrm{rank}(F_-)\le\min(N,M)$$
and
$$\mathrm{rank}(F_+)\le\min(N,M)$$

These four inequalities only allow for one solution, which is $M=N$. To see this, assume that for example $N&lt;M$, then it follows that $\mathrm{rank}(F_-)=\mathrm{rank}(F_+)\leq N$ but also $\mathrm{rank}(F_+)\geq M$, from which $N\geq M$ would follow, making the assumption inconsistent. The same goes for assuming $N&gt;M$.

That means the discrete Fourier transform must be between complex vector spaces of equal dimensions. Or in other words, the number of frequency bins must be equal to the number of time samples.

The proof above can of course be a single line if you already know that vector space isomorphisms require matching dimensions. Of course, some linear algebra is still necessary, but the rank formulas should be rather intuitive.