I believe that we can prove everything using the fact that Sylow $p$-subgroups are maximal $p$-subgroups (by definition).
Let $G = H \times K$ and let $P$ be a Sylow $p$-subgroup of $G$. Consider the projections onto each coordinate of the direct product:
\begin{align}
\pi_1 : H \times K &\rightarrow H, \quad
\pi_2 : H \times K \rightarrow K
\\
(h,k) &\mapsto h, \quad
(h,k) \mapsto k
\end{align}
Notice that $P \leq \pi_1(P) \times \pi_2(P) \leq G$. But $\pi_1(P) \times \pi_2(P)$ is a $p$-group (see note below) and so since $P$ is a Sylow $p$-subgroup (and thus is maximal), we must have $P = \pi_1(P) \times \pi_2(P)$.
Now let's show that $\pi_1(P)$ is a Sylow $p$-subgroup of $H$ and $\pi_2(P)$ is a Sylow $p$-subgroup of $K$.
Without loss of generality, suppose $\pi_1(P)$ is not a Sylow $p$-subgroup of $H$. Since $\pi_1(P)$ is a $p$-subgroup (see note below), then in particular our assumption implies that it is not maximal in $H$. Then there exists a $p$-subgroup $Q \leq H$ with $|\pi_1(P)| < |Q|$. But then $Q \times \pi_2(P)$ is a $p$-subgroup of $G$ with
\begin{align}
|P| = |\pi_1(P) \times \pi_2(P)| < |Q \times \pi_2(P)|
\end{align}
But this is impossible, because $P$ is maximal.
Note: I have used the following general property of elements of direct products:
\begin{align}
|(h,k)| = \text{lcm}(|h|,|k|)
\end{align}
So in particular, if $|(h,k)| = p^n$ for some $n$, then we know that both $h$ and $k$ must have order a power of $p$.