Prove $\lim(s_n \cdot t_n) = (\lim s_n)\cdot(\lim t_n)$

Fact 1: A convergent sequence of the number real is bounded.

Fact 2: Let $\lbrace s_n\rbrace\rightarrow s$ so, if $\varepsilon>0$ there existe $N_1>0$, such that for all $n\geq N_1$, implies $|s_n-s|<\frac{\varepsilon}{2M}$.

Fact 3: Let $\lbrace t_n\rbrace\rightarrow t$ so, if $\varepsilon>0$ there existe $N_2>0$, such that for all $n\geq N_2$, implies $|t_n-t|<\frac{\varepsilon}{2M}$.

Now

\begin{align} |s_nt_n-st|&\leq |t_n||s_n-s|+|s||t_n-t|\\ \end{align} Since $\lbrace t_n\rbrace$ is bounded, there exist $K>0$, such that $|t_{n}|\leq K$, and let $M:=Sup\lbrace K,|s|\rbrace$, now: \begin{align} |s_nt_n-st|&\leq |t_n||s_n-s|+|s||t_n-t|\\ &< M\frac{\varepsilon}{2M}+M\frac{\varepsilon}{2M}=\varepsilon. \end{align} For all $n>max\lbrace N_1,N_2\rbrace$