You can bound $\lvert N(S)\rvert$ from below by considering the worst case that all neighbours have the maximal degree, $n-d+1$. Then each vertex in $S$ of degree $n-d+1$ has as many outgoing edges as a worst-case neighbour has incoming edges, so the "cost balance" of such a vertex is neutral. Each vertex in $S$ of degree $n-d$ "saves" one edge, but there can be at most $n-d$ of these in $S$, so only $n-d$ edges can be "saved", whereas $n-d+1$ edges would have to be "saved" to "save" one worst-case neighbour. So no neighbour can be "saved" even in the worst case, and thus there must be as many neighbours as vertices in $S$.
I can write that out in formulas if you want, but you said you wanted a hint.
[Edit in response to request for formulas:]
Since all edges incident at a vertex in $S$ are also incident at a vertex of $N(S)$, we have
$$\sum_{w\in N(S)}\deg(w)\ge\sum_{v\in S}\deg(v)\;.$$
If $S$ contains $n_-$ vertices of degree $n-d$ and $n_+$ vertices of degree $n-d+1$, then
$$\begin{eqnarray}
\sum_{v\in S}\deg(v)&=&n_-(n-d)+n_+(n-d+1)\\\
&=&(n_-+n_+)(n-d+1)-n_-\\\
&=&\lvert S \rvert (n-d+1)-n_-\\\
&\ge&\lvert S \rvert (n-d+1)-(n-d)\;,
\end{eqnarray}$$
since there are only $n-d$ vertices of degree $n-d$. On the other hand, since no vertex has degree more than $n-d+1$, we also have
$$\sum_{w\in N(S)}\deg(w)\le \lvert N(S) \rvert (n-d+1)\;.$$
Putting this all together gives
$$\lvert N(S) \rvert (n-d+1) \ge \lvert S \rvert (n-d+1)-(n-d)\;,$$
$$\lvert N(S) \rvert \ge \lvert S \rvert -\frac{n-d}{n-d+1}\;.$$
But $\lvert N(S) \rvert$ and $\lvert S \rvert$ are integers and the last term is less than $1$, so we can drop the last term to conclude that $\lvert N(S) \rvert\ge\lvert S \rvert$. Linking this back to the original answer, each vertex in $S$ of degree $n-d$ can change the balance in the penultimate inequality by at most $1$, and there are at most $n-d$ such vertices, so the balance can only be changed by $n-d$, but it would have to change by $n-d+1$ to make a difference in the last inequality, where the division by $n-d+1$ corresponds to the fact that we need to "save" $n-d+1$ edges in order to "save" one worst-case neighbour.
If a bipartite graph has parts of size $m$ and $n$, it has at most $mn$ edges.
What are the possibilities if $m + n = 17$? In particular, how large can $mn$ be?
Best Answer
I don't think what you have said is incorrect, but maybe a little more involved than is necessary. Call the degree-9 vertex $v$. There are at least nine vertices opposite $v$. So there are at most 3 vertices on the side that $v$ is on. So there must be at least one (actually at least four) degree-6 vertex on the side opposite $v$. But this means there are at least 6 vertices on the same side $v$. "at most 3" contradicts "at least 6".