All the root spaces ${\U0001d524}_{\alpha}$ are 1dimensional.
Root spaces are 1dimensional
Recap
If $\U0001d524$ is the complexified Lie algebra of a compact semisimple group $K$ then $$\U0001d524=\U0001d525\oplus \underset{\alpha \in R}{\oplus}{\U0001d524}_{\alpha}$$ where:

$\U0001d525$ is the complexified Lie algebra of a maximal torus $T\subset K$ ,

${\U0001d524}_{\alpha}=\{X\in \U0001d524:{\mathrm{ad}}_{H}X=\alpha (H)X\forall H\in \U0001d525\}$ is the root space corresponding to the root $\alpha \in {\U0001d525}_{\mathbf{Z}}^{*}\subset {\U0001d525}_{\mathbf{R}}^{*}$ .
The figure below shows the root diagram of $\U0001d530\U0001d529(3,\mathbf{C})$ . 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 $\alpha $ .
If $X\in {\U0001d524}_{\alpha}$ and $Y\in {\U0001d524}_{\alpha}$ then $X$ , $Y$ and ${H}_{\alpha}=[X,Y]$ span a subalgebra ${S}_{\alpha}\cong \U0001d530\U0001d529(2,\mathbf{C})$ .
Partway through the proof of the previous point, we saw that $[X,Y]=K(X,Y){\alpha}^{\mathrm{\u266f}}\in \U0001d525$ , where ${\alpha}^{\mathrm{\u266f}}$ 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 $K(X,Y)=2/K({\alpha}^{\mathrm{\u266f}},{\alpha}^{\mathrm{\u266f}})$ ) then ${H}_{\alpha}=[X,Y]=\frac{2{\alpha}^{\mathrm{\u266f}}}{K({\alpha}^{\mathrm{\u266f}},{\alpha}^{\mathrm{\u266f}})}$ satisfies the commutation relations $$[{H}_{\alpha},X]=2X,[{H}_{\alpha},Y]=2Y,[X,Y]={H}_{\alpha}.$$
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 ${\U0001d524}_{\alpha}$ , ${\U0001d524}_{2\alpha}$ , etc and ${\U0001d524}_{\alpha}$ , ${\U0001d524}_{2\alpha}$ etc, that is root spaces corresponding to integer multiples of $\alpha $ .
The result is a subspace $$V=\mathbf{C}\cdot {H}_{\alpha}\oplus \underset{k\in \mathbf{Z}\setminus \{0\}}{\oplus}{\U0001d524}_{k\alpha}$$
$V$ 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 ${\U0001d524}_{k\alpha}$ are all 1dimensional.
We need to show that ${\mathrm{ad}}_{X}$ , ${\mathrm{ad}}_{Y}$ and ${\mathrm{ad}}_{{H}_{\alpha}}$ preserve $V$ . Let's just do ${\mathrm{ad}}_{X}$ (the others are similar/easier).

We have ${\mathrm{ad}}_{X}{H}_{\alpha}=[X,{H}_{\alpha}]=[{H}_{\alpha},X]=2X\in {\U0001d524}_{\alpha}\subset V$ , so ${\mathrm{ad}}_{X}$ sends $\mathbf{C}\cdot {H}_{\alpha}$ to something in $V$ .

We have ${\mathrm{ad}}_{X}:{\U0001d524}_{k\alpha}\to {\U0001d524}_{(k+1)\alpha}$ because $X\in {\U0001d524}_{\alpha}$ .

If $k+1\ne 0$ we have ${\U0001d524}_{(k+1)\alpha}\subset V$ and we're done.

If $k+1=0$ then ${\U0001d524}_{(k+1)\alpha}={\U0001d524}_{0}$ , which is not contained in $V$ . In fact, we have $V\cap {\U0001d524}_{0}=\mathbf{C}\cdot {H}_{\alpha}$ , so we need to show that ${\mathrm{ad}}_{X}Y$ is a multiple of ${H}_{\alpha}$ for any $Y\in {\U0001d524}_{\alpha}$ . But we saw that if $X\in {\U0001d524}_{\alpha}$ and $Y\in {\U0001d524}_{\alpha}$ then $[X,Y]=K(X,Y){\alpha}^{\mathrm{\u266f}}$ , and ${\alpha}^{\mathrm{\u266f}}$ is just a multiple of ${H}_{\alpha}$ , so ${\mathrm{ad}}_{X}({\U0001d524}_{\alpha})=\mathbf{C}\cdot {H}_{\alpha}\subset V$ 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 ${\U0001d524}_{k\alpha}$ is a weight space with weight $2k$ for the action of ${H}_{\alpha}$ .
This is because if $Z\in {\U0001d524}_{k\alpha}$ , we have $${\mathrm{ad}}_{{H}_{\alpha}}Z=k\alpha ({H}_{\alpha})Z$$ and, since ${H}_{\alpha}=2{\alpha}^{\mathrm{\u266f}}/K({\alpha}^{\mathrm{\u266f}},{\alpha}^{\mathrm{\u266f}})$ , we have $\alpha ({H}_{\alpha})=2\alpha ({\alpha}^{\mathrm{\u266f}})/K({\alpha}^{\mathrm{\u266f}},{\alpha}^{\mathrm{\u266f}})$ , but $\alpha ({\alpha}^{\mathrm{\u266f}})=K({\alpha}^{\mathrm{\u266f}},{\alpha}^{\mathrm{\u266f}})$ , so $${\mathrm{ad}}_{{H}_{\alpha}}Z=2kZ$$ and $Z$ has weight $2k$ under the action of ${H}_{\alpha}$ as required.
So our weight diagram looks like this: $$\mathrm{\cdots}\oplus {\U0001d524}_{2\alpha}\oplus {\U0001d524}_{\alpha}\oplus \mathbf{C}\cdot {H}_{\alpha}\oplus {\U0001d524}_{\alpha}\oplus {\U0001d524}_{2\alpha}\oplus \mathrm{\cdots}$$ in weights $$\mathrm{\cdots}\mathit{\hspace{1em}}4,2,0,2,4,\mathrm{\cdots}$$ respectively. In particular, the weight space with weight zero is $\mathbf{C}\cdot {H}_{\alpha}$ , 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.