Let $G$ be a compact connected semisimple Lie group and let $\mathfrak g$ denote its Lie algebra. 
Is the following result true? Does it follows directly from Dynkin's classification of maximal Lie subalgebras?:



> 
> There are $\mathfrak h\_1,\dots,\mathfrak h\_r$ proper Lie subalgebras of $\mathfrak g$ such that: (1) for any Lie subalgebra $\mathfrak a$ of $\mathfrak g$ there is $a\in G$ such that $\textrm{Ad}\_a(\mathfrak a)\subset \mathfrak h\_i$ for some $i$;
>  (2) if a Lie subalgebra $\mathfrak a$ of $\mathfrak g$ contains properly $\mathfrak h\_i$ for some $i$, then $\mathfrak a=\mathfrak g$;
>  (3) for each $i$, the only connected Lie subgroup $H\_i$ of $G$ with Lie algebra $\mathfrak h\_i$ is closed in $G$. 
> 
> 
> 


I am confident that the answer is affirmative, except in the finiteness of the number of Lie subalgebras $\mathfrak h\_i$. 
I hope there is a precise (modern) reference for it.


Some related questions:


[Modern reference for maximal connected subgroups of compact Lie groups](https://mathoverflow.net/q/60315) 


[Maximal subgroups of semisimple Lie groups](https://mathoverflow.net/q/127352)



On maximal closed connected subgroups of a compact connected semisimple Lie group?
It's true, and does not rely on classification of maximal subalgebras.


The first part of the statement is that there are finitely maximal subalgebras up to conjugation. It follows from the fact that in a semisimple real Lie algebra there are finitely many semisimple subalgebras up to conjugation, and that every maximal subalgebra in a compact semisimple real Lie algebra, if not abelian (modulo its core, see edit), is product of a semisimple subalgebra with its centralizer, and if abelian (modulo its core) has to be a maximal torus (in the quotient by the core) and there's only a single such conjugacy class.


For being closed: if $\mathfrak{h}$ is a maximal subalgebra and $H$ its corresponding analytic subgroup, I claim that $H$ is closed. Otherwise, $H$ is dense (since otherwise its Lie algebra would be larger), and hence $\mathfrak{h}$ is $G$-invariant, hence an ideal, hence not maximal.




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*Edit:* Recall that the core is the largest ideal in the subalgebra. There are finitely many possibilities, since any semisimple Lie algebra has finitely many ideals. Actually, even if I didn't use it, it is not hard to prove that the quotient by core of a maximal subalgebra in a semisimple Lie algebra 
is either simple, or direct product of two isomorphic simple Lie algebras, in such a way that the maximal subalgebra modulo its core is the graph of an isomorphism between the factors. Given this, everything boils down to the case when $\mathfrak{g}$ is simple.


