 Solution

For the standard A4 root system inside the A5 root system, let $p$ delete the fifth Dynkin label. Put $v_i=p(w_i)$ for $i\le 5$. These are precisely the five weights of the defining $SU(5)$ fundamental representation:Since $p(w_6)=0$, the six projected weights of $\overline{6}$ are $-v_1,\dots ,-v_5,0$. The first five form the dual representation $\overline{5}$, with highest weight $(0,0,0,1)$; the last is a singlet. ThusFor $15$, split the fifteen weights in b into the ten pairs with $1\le i<j\le 5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete weight decomposition of ${\Lambda }^{2}5=10$, with highest weight $(0,1,0,0)$. The latter project to $v_i$, the complete weight decomposition of $5$. ThereforeThis restriction of a representation can also be checked without a table: $V{\downarrow }_{SU(5)}=5\oplus 1$, so its dual representation is $\overline{5}\oplus 1$, while its exterior square is ${\Lambda }^{2}5\oplus (5\otimes 1)$. The dimensions match $6=5+1$ and $15=10+5$.