If we have two Dynkin diagrams such that $D_{1}$ is a subdiagram of $D_{2}$. Then the $\mathfrak{g}(R_{1})$ is a subalgebra of $\mathfrak{g}(R_{2})$

Question

Recently I was studying my notes on Lie algebra, and while I was studying Dynkin diagrams, I came with the following question:

If we have $D_{1}$ and $D_{2}$ two Dynkin diagrams such that $D_{1}$ is a subdiagram of $D_{2}$, and $R_{1}$ and $R_{2}$ are the root systems corresponding to $D_{1}$ and $D_{2}$, respectively. Then $\mathfrak{g}(R_{1})$ is a subalgebra of $\mathfrak{g}(R_{2})$.

So here I have some questions:

1. Is it sufficient to have that $D_{1}$ is a subdiagram of $D_{2}$ to then have that $\mathfrak{g}(R_{1})$ is a subalgebra of $\mathfrak{g}(R_{2})$? why would this be true?

2. If it is not sufficient, what else do we need?

3. Is this a famous result? I feel it could be a result in a book or somewhere, but I haven't found anything similar (maybe I am bad at searching).

Regarding my questions, for 1. Using my intuition I think the answer is yes, but I don't know how to prove it, so any help in that regard would be appreciated. For 2. I really cant come up with more conditions, but the fact that I haven't been able to prove the arguments makes me doubt.