I think you can pull an argument like this : define the sequence of compacts $[-n, n]$ ; since $\mathcal A$ is dense in each of them, for each function $f \in L^p(\mathbb R)$ and for all $n$, there exists a sequence $f_{n,m}(x)$ such that $f_{n,m} \to f$ in $[-n,n]$ with respect to the norm you've chosen as $m \to \infty$.
You could extract a subsequence out of the sequences $f_{n,m}$ by the following argument. Choose a sequence $\varepsilon_k \to 0$. We will find $g_k$ such that
$$
\|g_k - f\|_{\infty} \overset{def}{=} \inf \{ K > 0 \, | \, \mu(\{x \in \mathbb R \, | \, |g_k(x) - f(x)| > K\} ) = 0 \} < \varepsilon_k.
$$
(I believe this is the detailed definition of your norm? Correct me if I'm wrong.)
with $\mu$ being the Lebesgue measure.
Now $f \in L^p$, so that for $n$ big enough, $|f(x)| < \varepsilon_k/2$ almost everywhere outside $[-n,n]$. Fix this $n$.
This is the point where I am stuck at. Since $f_{n,m} \to f$ in $[-n,n]$, there exists $m$ such that $\|f_{n,m}-f \|_{\infty} < \varepsilon_k$ inside $[-n,n]$ and $|f_{n,m}(x)| < \varepsilon_k/2$ outside$^{*}$ $[-n,n]$ ($f_{n,m} \in L^p$ also, hence goes to $0$ at infinity.)
This guy ($^{*}$) is the problem ; I don't have any control over what happens to $f_{n,m}$ outside $[-n,n]$ and that is what gives me intuition that it might be false in the general case. Although your choice of functions exhibit more properties than the abstract one... but this is now at a state of intuition only. Let me finish just to see where I froze in my argument.
With those properties summed up, outside $[-n,n]$, $|f_{n,m}(x) - f(x)| \le |f_{n,m}(x)| + |f(x)| < \varepsilon_k$ almost everywhere outside $[-n,n]$ and $|f_{n,m}(x) - f(x)| < \varepsilon_k$ inside $[-n,n]$, so that $|f_{n,m}(x)-f| < \varepsilon_k$ almost everywhere over $\mathbb R$, that is, we have found $f_{n,m}$ such that $\|f_{n,m} - f\|_{\infty} < \varepsilon_k$. Just let $g_k = f_{n,m}$.
I hope it helped in some way to at least give you some ideas.
EDIT : I forgot the "$ = 0$" in my definition of the norm. Typo error.
Best Answer
If $(x_1,y_1) \neq (x_2,y_2)$ the either $x_1 \neq x_2$ or $y_1 \neq y_2$. In the first case there exists $f \in C(X)$ such that $f(x_1)\neq f(x_2)$. The function $h(x,y)=f(x)$ is in $\mathcal A$ and $h(x_1,y_1) \neq h(x_2,y_2)$. The second case is similar, so $\mathcal A$ separates points. Rest of the argument you have given is fine.