Let $x_n\to x$ and $y_n\to y$ in the space $\Bbb{R}$. Then $x_n +y_n \to x+y$, $x_n -y_n\to x-y$, $x_n \cdot y_n\to x\cdot y$, and provided that each $y_n\neq 0$ and $y\neq 0$, $x_n /y_n \to x/y$.
My attempt:
Approach(1):
Claim: Assume $X$ is metrizable. $\{a_n \}\to a$ convergence in metric space notion $\iff$ $\{a_n \}\to a$ in topological sense. In other word, $\forall \epsilon \gt 0$ $\exists N\in \Bbb{N}$ such that $d(a,a_n)\lt \epsilon ,\forall n\geq N$ $\iff$ $\forall U\in \mathcal{N}_a$ $\exists N\in \Bbb{N}$ such that $a_n\in U, \forall n\geq N$.
Proof: ($\Rightarrow$) let $U\in \mathcal{N}_a$. Since $U\in \mathcal{T}_d=\mathcal{T}_X$ And $a\in U$, $\exists \epsilon \gt 0$ such that $B_d(a,\epsilon)\subseteq U$. For $\epsilon \gt 0, \exists N\in \Bbb{N}$ such that $d(a,a_n)\lt \epsilon ,\forall n\geq N$. Which implies $a_n\in B_d(a,\epsilon),\forall n\geq N$. Thus $a_n\in U, \forall n\geq N$. ($\Leftarrow$) conversely let $\epsilon \gt 0$. Clearly $B_d(a,\epsilon)\in \mathcal{N}_a$. Then $\exists N\in \Bbb{N}$ such that $a_n\in B_d(a,\epsilon),\forall n\geq N$. Which implies $d(a,a_n)\lt \epsilon, \forall n\geq N$. This completes the proof.
$\Bbb{R}$ is metrizable with euclidean norm. By theorem 3.3 of baby Rudin and above claim, all the sequence given in problem convergence. Is this proof correct?
Approach(2): (following hint) $\Bbb{R}^2$ equipped with $\mathcal{T}_p$(product topology)=$\mathcal{T}_b$(box topology). By exercise 6 section 19, since $x_n \to x$ and $y_n \to y$, we have $x_n \times y_n \to x\times y$. By lemma 21.4 or exercise 12 section 21, $+$, $-$, $\cdot$ and $/$ maps are continuous. By theorem 21.3, $\{f(x_n\times y_n)\}_{n\in \Bbb{N}}\to f(x\times y)$, where $f$ is $+$, $-$, $\cdot$ and $/$ map. Is this proof correct?
Edit: Domain for $/$ map is $\Bbb{R}\times (\Bbb{R}\setminus \{0\} )$ with product topology. Set $\Bbb{R}\setminus \{0\}$ equipped with subspace topology. So we can apply exercise 6 section 19.