Let $F$ be the filter of statistical convergence on $\mathbb{N}$ (see Definition 2.1 here). Suppose that $(\xi_n), (\eta_n)$ are sequences of real numbers that converge to $\xi$ and $\eta$ along $F$, respectively. Can we prove directly that $(\xi_n\cdot \eta_n)$ converges to $\xi\cdot \eta$ along $F$?
It is easy to prove a similar fact for ultrafilters. So the idea is to extend $F$ to an ultrafilter and go back to $F$ but is it necessary?
If you already know that a sequence $(x_n)$ is statistically convergent to $x$ if and only if there exists a set $A$ such that $d(A)=1$ and the subsequence $(x_n)_{n\in A}$ is convergent, then the proof is almost immediate: Take the subset $A$ for one sequence, a set $B$ for the other one. Then the subsequence $(x_ny_n)_{n\in A\cap B}$ converges in the usual sense. Since $d(A\cap B)=1$, this implies statistical convergence.
The symbol $d(A)$ denotes the asymptotic density of the set $A$. The proof of the fact I have mentioned above can be found, for example, in this paper: Šalát, Tibor, On statistically convergent sequences of real numbers. (English). Mathematica Slovaca, vol. 30 (1980), issue 2, pp. 139-150, http://dml.cz/handle/10338.dmlcz/136236
However, it is not difficult to prove this even without using the aforementioned results and in a somewhat greater generality.
Let $\newcommand{\FF}{\mathcal F}\FF$ be any filter on the set $\mathbb N$ and let $(x_n)$, $(y_n)$ be sequences such that $\newcommand{\Flim}{\operatorname{\FF-lim}}\Flim x_n=x$ and $\Flim y_n=y$. Then $\Flim (x_n y_n) = xy$.
Here $\Flim x_n=x$ means that for each $\varepsilon>0$ $$\{n\in\mathbb N; |x_n-x|\le\varepsilon\}\in\FF.$$ For more on this type of convergence see here (and some of the questions linked there) or here (and some of the questions linked there).
The claim above has, as a special case, the claim for statistical convergence. (You just choose the filter of all sets having asymptotic density equal to $1$.) The proof is basically the same as for the usual limit.
Proof. Let $\newcommand{\ve}{\varepsilon}\ve>0$. Then there exists a set $F_1\in\FF$ such that $$(\forall n\in F_1) |x_n-x|<\ve.$$ Moreover, we can also choose $F_1$ in such way that $|x_n-x|\cdot |y|<\ve$. (If $y=0$, this condition is true. If not, we take $|x_n-x|<\min\{\ve,\frac{\ve}{|y|}\}$.)
Let $M=|x|+\ve$. Then for every $n\in F_1$ we have $|x_n|<M$.
There exists $F_2$ such that $$(\forall n\in F_2) |y_n-y|<\frac{\ve}M.$$ The we get for $n\in F:=F_1\cap F_2$ $$|x_ny_n-xy| = |x_n(y_n-y)+(x_n-x)y| \le |x_n| \cdot |y_n-y| + |x_n-x| \cdot |y| < M \cdot \frac{\ve}M + \ve = 2\ve.$$ The set $F$ belongs to the filter $\FF$.
Thus the sequence $(x_ny_n)$ is $\FF$-convergent to $xy$. $\hspace{5cm}\square$
The proof of this fact is also given, for example, in this paper. (Although it is formulated there using ideals rather then filters.)
