If H is a subgroup of G, then H has no more Sylow subgroups than G 
If $H$ is a subgroup of the finite group $G$, then how do I show that $n_p(H) \leq n_p(G)$? Here $n_p(X)$ is the number of Sylow $p$-subgroups in the finite group $X$.

Here is my attempt: Suppose the order of $G$ is $n$, the order of $H$ is $m$, then $m$ divides $n$ by Lagrange's theorem. By Sylow's counting theorem $n_p(H) = 1 \pmod p$ and $n_p(H)$ divides $m$, $n_p(G) =1 \pmod p$, $n_p(G)$ divides $n$. 
How do I continue after this? 
 A: If $Q \in \operatorname{Syl}_p(H)$, then $Q \leq P$ for some $P \in \operatorname{Syl}_p(G)$. Then $Q \leq P \cap H \leq H$ and $P \cap H$ is a $p$-group,  so it follows that $Q = P \cap H$.
In other words, we have the inclusion $\operatorname{Syl}_p(H) \subseteq \{P \cap H: P \in \operatorname{Syl}_p(G) \}$ which gives the result. This inclusion is an equality when $H$ is a normal subgroup, but not in general.
A: This is really a consequence of the pigeonhole principle: Assume that $H$ has more Sylow $p$-subgroups than $G$. Since every Sylow $p$-subgroup of $H$ is contained in a Sylow $p$-subgroup of $G$, there exists a Sylow $p$-subgroup $P$ of $G$ containing two distinct Sylow $p$-subgroups $Q_1$ and $Q_2$ of $H$. Since $Q_1$ and $Q_2$ are distinct, the group generated by them is not a $p$-group; on the other hand, $<Q_1,Q_2>\le P$ which is a $p$-group, a contradiction.
A: Bonus information: there is a paper of prof Gabriel Navarro from 2003 that implies that for a solvable group $G$, and any prime $p$ dividing $|G|$, $n_p(H) \mid n_p(G)$ for every subgroup $H$ of $G$. Here $n_p(X)=\#Syl_p(X)$ (which is understood to be $1$ if $p \nmid |X|$).
A: I like this question, and wanted it to have a bit of a longer answer:
Surprising
This result should be a little surprising. After all the Sylow $p$-subgroups of $H$ can have smaller order, and even though $G$ may have only a few subgroups of order $p^n$, it might have a ton of order $p^{n-1}$. For example, in $G=A_4$, we have only $n_2(G)=1$ Sylow $2$-subgroup, but it has $3$ subgroups of order $2^1$, and so a subgroup $H$ has $n_2(H) \leq 3$, but that leaves open the possibility of $n_2(H) \in \{2,3\}$ both of which contradict the theorem. There is an analogue of $A_4$ called $G=\operatorname{AGL}(1,p^2)$ with approximately the same behavior: $n_p(G)=1$ but $G$ has $p+1$ subgroups of order $p$. When one actually looks for a subgroup $H$, one runs into a problem: no single $H$ contains all those subgroups of order $p$ (or even two of them in the $A_4$ and AGL cases) unless it contains an entire Sylow $p$-subgroup (making those smaller $p$-subgroups irrelevant).
Proof
This idea leads to a simple proof (given by Derek Holt in the comments).
We construction a 1-1 function $f$ from $\operatorname{Syl}_p(H)$ to $\operatorname{Syl}_p(G)$ where $\operatorname{Syl}_p(X)$ is the set of Sylow $p$-subgroups of the finite group $X$. Given a Sylow $p$-subgroup $Q$ of $H$, Sylow's containment theorem says that $Q$ (a $p$-subgroup of $G$) is contained in some Sylow $p$-subgroup $f(Q)$ of $G$. If $f(Q_1) = f(Q_2)$, then $Q_1, Q_2 \leq f(Q_1)$ and $Q_1, Q_2 \leq H$, so $Q_1, Q_2 \leq f(Q_1) \cap H$. However, $f(Q_1) \cap H$ is a $p$-subgroup of $H$, and a $p$-subgroup of $H$ can only contain at most a single Sylow $p$-subgroup of $H$, so $Q_1 = Q_2$. Hence $f$ is a 1-1 function, and $n_p(H) \leq n_p(G)$.
Normal subgroups
If $H$ is a normal subgroup of $G$, then in fact $n_p(H)$ divides $n_p(G)$. Hall (1967) calculated $n_p(G) = n_p(H) \cdot n_p(G/H) \cdot n_p(T)$ where $T=N_{PH}(P \cap H)$ for any Sylow $p$-subgroup $P$ of $G$.
(See also this answer of Mikko Korhonen.)
