I just began to learn Analysis by myself. Can someone please help me check my proof to see any errors in math, logic and English. Thank you so much!!
Fact: Let $X$ be a topological space. Let $S\subseteq X$ and $x\in X$. Then $x\in\text{cl}~S$ (the closure of $S$) if and only if there is a net $(x_\lambda)$ in $S$ such that $x_\lambda\rightarrow x$.
Theorem: Let $X$ be a normed space and $A$ be a non-empty countable subset of $X$. Then $\overline{\operatorname{span}}(A)$ is separable.
Proof: Let $Q_0$ be the rationals (if $\mathbb{F}=\mathbb{R}$) or the complex numbers with rational real and imaginary parts (if $\mathbb{F}=\mathbb{C}$). Then $Q_0$ is dense in $\mathbb{F}$. Let $$S=\{\sum\limits_{i=1}^{n}q_ix_i:n\in\mathbb{N},q_i\in Q_0,x_i\in A\}.$$ Since $Q_0$ and $A$ are countable, $Q_0^n\times A^n$ is countable for any $n\in\mathbb{N}$. For any $n\in\mathbb{N}$, let $S_n=\{\sum\limits_{i=1}^{n}q_ix_i:q_i\in Q_0,x_i\in A\}$ and let $f:Q_0^n\times A^n\rightarrow S_n$, $(q_1,\cdots,q_n,x_1,\cdots,x_n)\mapsto\sum\limits_{i=1}^{n}q_ix_i$. Then $S_n$ is countable for any $n\in\mathbb{N}$ as a function image of a countable set is also countable. Since $S=\bigcup\limits_{n\in\mathbb{N}}S_n$, $S$ is a countable subset of $\text{span}(A)$. Let $x\in\text{span}(A)$. Then $x=\alpha_1x_1+\cdots+\alpha_mx_m$ for some $m\in\mathbb{N}$, $x_1,\cdots$, $x_m\in A$ and $\alpha_1,\cdots$, $\alpha_m\in\mathbb{F}$. Since $Q_0$ is dense in $\mathbb{F}$, $\alpha_1,\cdots$, $\alpha_m\in\mathbb{F}=\text{cl}~Q_0$. By the Fact above, there exists a sequence $(\alpha_{j,k})_{k=1}^{\infty}$ in $Q_0$ such that $\alpha_{j,k}\rightarrow\alpha_j$ for each $j$ ($j=1,\cdots$, $m$). It follows from the continuity of the linear space operations that there exists a sequence $(\alpha_{1,k}x_1+\cdots+\alpha_{m,k}x_m)_{k=1}^{\infty}$ in $S$ such that $\alpha_{1,k}x_1+\cdots+\alpha_{m,k}x_m\rightarrow\alpha_1x_1+\cdots+\alpha_mx_m$. By the Fact above, $x\in clS$. It implies that $\text{span}(A)\subseteq\text{cl}~S$. Thus $S$ is dense in $\text{span}(A)$. Since the closed linear span contains the linear span, $S$ is dense in $\overline{\operatorname{span}}(A)$. Therefore, $\overline{\operatorname{span}}(A)$ is separable.