All the root spaces little g alpha are 1dimensional.
Root spaces are 1dimensional
Recap
If little g is the complexified Lie algebra of a compact semisimple group K then little g splits little h direct sum a direct sum over roots alpha of subspaces little g alpha where:

little h is the complexified Lie algebra of a maximal torus T\subset K,

little g alpha equals the set of X in little g such that little ad_H X equals alpha of H times X for all H in little h is the root space corresponding to the root alpha in little h Z star inside little h R star.
The figure below shows the root diagram of little s l 3 C. We've been gradually discovering that various nice features of this Lie algebra and its root diagram carry over to root diagrams in this more general context. These nice features include:
If alpha is a root then so is minus alpha.
If X is in little g alpha and Y is in little g minus alpha then X, Y and H_alpha equals X bracket Y span a subalgebra S alpha isomorphic to little s l 2 C.
Partway through the proof of the previous point, we saw that X bracket Y equals K of X, Y times alpha sharp in little h, where alpha sharp is Killingdual to alpha. This depends on X and Y only through their Killing pairing K(X, Y). We also saw that if we pick X and Y suitably (so that K of X, Y equals 2 over K of alpha sharp, alpha sharp) then H alpha equals X bracket Y equals 2 alpha sharp over K alpha sharp, alpha sharp satisfies the commutation relations H alpha bracket X equals 2 X, H alpha bracket Y equals minus 2 Y, and X bracket Y equals H alpha.
In this video, we will show that:
Proof
Fix a root alpha and consider the line through alpha. We will sum some of the root spaces along this line:

over 0, we take the span of H_alpha.

we take the root spaces little g alpha, little g 2 alpha, etc and little g minus alpha, little g minus 2 alpha etc, that is root spaces corresponding to integer multiples of alpha.
The result is a subspace V equals the span of H_alpha direct sum the direct sum of little g k alphas where k runs over the nonzero integers.
V is preserved by the action of the subalgebra S alpha. Moreover, it is irreducible as a representation of S alpha.
Irreducibility is the key thing, because this implies that the root spaces little g k alpha are all 1dimensional.
We need to show that little ad_X, little ad_Y and little ad H alpha preserve V. Let's just do little ad_X (the others are similar/easier).

We have little ad_X H alpha equals X bracket H alpha, which equals minus H alpha bracket X, which equals minus 2 X, which is in little g alpha, which is contained in V, so little ad_X sends the span of H alpha to something in V.

We have little ad_X from little g k alpha to little g (k plus 1) times alpha because X is in little g alpha.

If k + 1 is nonzero we have little g (k + 1) times alpha is contained in V and we're done.

If k + 1 = 0 then little g (k + 1) alpha equals little g_0, which is not contained in V. In fact, we have V intersect little g_0 equals the span of H_alpha, so we need to show that little ad_X of Y is a multiple of H alpha for any Y is in little g minus alpha. But we saw that if X is in little g alpha and Y is in little g minus alpha then X bracket Y equals K of X, Y times alpha sharp, and alpha sharp is just a multiple of H alpha, so little ad_X applied to little g minus alpha is contained in the span of H_alpha, which is contained in V and we're done.

How do we see that this representation is irreducible? Let's understand the weight space decomposition of V under the action of S alpha. I claim that little g k alpha is a weight space with weight 2 k for the action of H alpha.
This is because if Z is in little g k alpha, we have little ad H alpha of Z equals k alpha of H alpha times Z and, since H alpha equals 2 alpha sharp over K alpha sharp, alpha sharp, we have alpha of H alpha equals 2 alpha of alpha sharp over K alpha sharp, alpha sharp, but alpha of alpha sharp equals K of alpha sharp, alpha sharp, so little ad H alpha of Z equals 2 k Z and Z has weight 2 k under the action of H alpha as required.
So our weight diagram looks like this: dot dot dot, little g minus 2 alpha, little g minus alpha, the span of H alpha, little g alpha, little g 2 alpha, etc in weights dot dot dot, minus 4, minus 2, 0, 2, 4 etc respectively. In particular, the weight space with weight zero is the span of H alpha, which is 1dimensional. If we decompose V into irreducibles then these irreducibles all have even highest weight (there are only even weights) and in particular they will all have 1dimensional weight space in weight zero. But there is only room for one such irreducible piece, so V must itself be irreducible.