Could someone check this proof? This is a particular exercise of Terry Tao's Analysis I. I'm not interested in that old proof based on bounded sequences.
Lemma 1. Let $\varepsilon, \delta $. If $x$ and $y$ are $\varepsilon$-close and $z$ and $w$ are $\delta$-close, then $xy$ and $zw$ are $(\varepsilon|z|) + \delta|x|) + \varepsilon\delta)$-close.
Proposition. Let $x = LIM_{n \rightarrow \infty} a_n$,$y = LIM_{n \rightarrow \infty} b_n$ , and $x' = LIM_{n \rightarrow \infty} a'_n$ be real numbers. Then $xy$ is also a real number.
Proof
We just have to show that $xy = LIM_{n \rightarrow \infty} a_nb_n$ is a Cauchy sequence.
Let $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ two Cauchy Sequences. So for any $\varepsilon/2 > 0$ there exists an $N_0$ such that $j, k > N_0$, $|a_j - a_k| < \varepsilon/2$ and for any $\varepsilon/2 > 0$ there exists an $N_1$ such that $j, j > N_1$, $|b_j - b_k| < \varepsilon/2$. Let $N = max\{N_0,N_1\}$.
Suppose we have for $j, k$ > N, $|a_j*b_j - a_k*b_k|$. Employing Lemma 1, $$|a_j*b_j - a_k*b_k| < (\varepsilon/2 |b_j| + \varepsilon/2 |a_j| + \varepsilon )$$.
Let's call $\delta = (\varepsilon/2 |b_j| + \varepsilon/2 |a_j| + \varepsilon )$. Then, for any $\delta$, we can find an $N$ such that for $j,k > N$, $|a_jb_j - a_kb_k| < \delta$. So that's also a Cauchy sequence.