= Solution
The <highest root> of the <D5 root system> in our numbering is $\theta=e_1+e_2=\alpha_1+2\alpha_2+2\alpha_3+\alpha_4+\alpha_5$. Adjoin $\alpha_0=-\theta$. The <Extended Dynkin diagram> has an extra bond $0-2$. Deleting node $2$ leaves the two isolated nodes $0,1$ and the chain $4-3-5$, hence the <root subsystem> $A_1\oplus A_1\oplus A_3$.
This is a <maximal Lie subalgebra>, not merely a full-rank <Lie subalgebra>. In the orthogonal realization it is $\mathfrak{so}(4)\oplus\mathfrak{so}(6)\subset\mathfrak{so}(10)$, acting on a $4+6$ decomposition. The remaining off-diagonal part transforms as the irreducible <tensor product representation> $(\mathbf4,\mathbf6)$ of dimension $24$. Any intermediate <Lie subalgebra> containing one nonzero vector in that complement must contain the entire complement, because this <representation> is <irreducible>, and hence all $6+15+24=45$ dimensions. This proves maximality.
To apply the supplied projection, write a <D5 root system> <weight> as the column $w=(a,b,c,d,e)^T$ of <Dynkin labels>. Then
$$
Pw=(c+d+e,\ c,\ a+b+c+e,\ b+c+d,\ -a-b-c-d)^T.
$$
For example, the <highest weight> $w=(0,0,0,0,1)^T$ maps to $(1,0;1,0,0)^T$, and $w=(-1,0,1,-1,0)^T$ maps to $(0,1;0,0,1)^T$. Applying the same expression to all sixteen <weights> in <4/b/solution> gives exactly
$$
\{(\epsilon,0;f):\epsilon=\pm1,\ f\in F\}
\ \cup\
\{(0,\epsilon;\bar f):\epsilon=\pm1,\ \bar f\in\overline F\},
$$
where $F$ and $\overline F$ are the four-element <weight systems> in <4/a/solution>. Each set has eight distinct <weights>, each of multiplicity one. Since an $A_1$ doublet has <weights> $\pm1$ and an $A_1$ singlet has <weight> zero, the <branching of a representation> is
$$
\boxed{\mathbf{16}\longrightarrow R=(\mathbf2,\mathbf1,\mathbf4)\oplus(\mathbf1,\mathbf2,\overline{\mathbf4}).}
$$
The <highest weights> of these two summands are $(1,0;1,0,0)$ and $(0,1;0,0,1)$, and $2\cdot1\cdot4+1\cdot2\cdot4=16$ checks the dimensions. Here $SU(2)\times SU(2)\times SU(4)$ is the convenient covering <group>; its image inside $\operatorname{Spin}(10)$ has a common $\mathbb Z_2$ central identification. Both summands are trivial on $(-I_2,-I_2,-I_4)$.
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