Generating set of lie algebra su(3) I am looking for (an example of) a minimal set of Gell-Mann matrices such that their closure under the Lie bracket is all of $\mathfrak{su}(3)$. By minimal I mean the set should be as small as possible. 
Even though this seems quite simple, I can't make much sense of any of the sources that I have consulted. For example, here I found that this set should only contain two elements. Any help is much appreciated. 
 A: Well, you know it is not a set of two generators from elementary I,U,V-spin considerations in the eightfold way. Below, you see by inspection a set of just three will suffice. Avoid taking your three from one of the numerous su(2) subalgefbras, of course!
So take the set $\lambda_1,\lambda_2,\lambda_4$, for the sake of argument.
Commuting them pairwise generates $\lambda_3,\lambda_7,\lambda_6$. Commuting the first and second set nets $\lambda_5,\lambda_8$. This is straightforward given the sparseness of the structure constants in this basis.
It is easier to see from the root diagram, or the meson octet, in physics. I hope you are not asking for isomorphisms among sufficient sets of threes. I gave you one.
A: Start with the two simple roots $\alpha_1$ and $\alpha_2$, corresponding to $E_{\alpha_1}$ and $E_{\alpha_2}$ respectively.  The commutator will give you the third root $\alpha_3=\alpha_1+\alpha_2: E_{\alpha_3}=[E_{\alpha_1},E_{\alpha_2}]$.  Take the negative of these (corresponding to transpose conjugate of the respective matrices).  The commutators of $[E_{\alpha_1},E_{-\alpha_1}]$ and $[E_{\alpha_2},E_{-\alpha_2}]$ will give two linearly independent Cartan elements.
