= Solution
Let index $0$ denote the first coordinate, which the embedded $SU(5)$ fixes, and let $a,b=1,\ldots,5$ denote the remaining coordinates. The <fundamental representation> and its <conjugate representation> immediately give
$$
\boxed{\mathbf6\downarrow SU(5)=\mathbf1\oplus\mathbf5,\qquad
\bar{\mathbf6}\downarrow SU(5)=\mathbf1\oplus\bar{\mathbf5}.}
$$
For a <symmetric tensor>, $S^{00}$ is a <trivial representation>, $S^{0a}=S^{a0}$ is a $\mathbf5$, and $S^{ab}$ is the fifteen-dimensional <symmetric square of a group representation> of $\mathbf5$. For an <antisymmetric tensor>, $A^{00}=0$, $A^{0a}=-A^{a0}$ is a $\mathbf5$, and $A^{ab}$ is the ten-dimensional <exterior power> $\Lambda^2\mathbf5$. Hence
$$
\boxed{\mathbf{21}\downarrow SU(5)=\mathbf{15}\oplus\mathbf5\oplus\mathbf1,\qquad
\mathbf{15}\downarrow SU(5)=\mathbf{10}\oplus\mathbf5.}
$$
Here the left-hand $\mathbf{15}$ is the <antisymmetric tensor> of $SU(6)$, while the right-hand $\mathbf{15}$ in the first equation is the <symmetric tensor> of $SU(5)$; equal dimensions alone do not identify <representations> of different <groups>.
Finally decompose the traceless block <matrix>
$$
M=\begin{pmatrix}a&b\\c&D\end{pmatrix},\qquad a+\operatorname{tr}D=0.
$$
The traceless part of $D$ is the <Adjoint representation> $\mathbf{24}$, the column $c$ is $\mathbf5$, the row $b$ is $\bar{\mathbf5}$, and the one remaining diagonal <trivial representation> is generated by $\operatorname{diag}(-5,1,1,1,1,1)$. Thus
$$
\boxed{\mathbf{35}\downarrow SU(5)=\mathbf{24}\oplus\mathbf5\oplus\bar{\mathbf5}\oplus\mathbf1.}
$$
All dimensions match: $6=1+5$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.
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