Continuous functions that coincide in dense set are equal Any corrections or feedback to the proof below will be highly appreciated.
Proposition. We have that $f\colon\mathbb{R}\to\mathbb{R}$ e $g\colon \mathbb{R}\to\mathbb{R}$ are continuous functions. Let $X_\alpha$ be a dense set in $\mathbb{R}$. If $f(x)=g(x)$ for all $x\in X_\alpha$, then $f=g$.
Proof. As $f$ and $g$ are continuous functions, we have for every point $x_0\in\mathbb{R}$ such that $\lim\limits_{n\to \infty}{f(x_n)}=f(x_0)$ for every sequence of points $(x_n)_n \in \Bbb R$ with $\lim\limits_{n\to\infty} x_n = x_0$, and analogous definition is valid for $g$.
Because $X_\alpha$ is dense, then $\overline{X_\alpha}=\mathbb{R}$, that is, we can find for every $z\in\mathbb{R}$ a sequence $(z_n)_n$ in $X_a$ such that $z_n\to z$.
Then for every $z\in\Bbb R$ we do exactly that, defining a sequence $(z_n)_n$ in $X_\alpha$ such that $z_n\to z$. And as $f$ and $g$ are both continuous, we say that we have $f(z_n)\to f(z)$ as well as $g(z_n)\to g(z)$.
And if $\lim\limits_{n\to\infty}{f(z_n)}=f(z)$ e $\lim\limits_{n\to\infty}{g(z_n)}=g(z)$, then we have that
$$\forall \epsilon_1>0\ \exists n_f\in\mathbb{N}:\ n>n_f\implies \left|f(z_n)-f(z)\right|<\epsilon_1$$
and also
$$\forall \epsilon_2>0\ \exists n_g\in\mathbb{N}:\ n>n_g\implies \left|g(z_n)-g(z)\right|<\epsilon_2$$
We take $n_0=\min{(n_f,n_g)}$, and we have
$$n>n_0\implies \left|f(z_n)-f(z)\right|<\epsilon_1\ \text{e}\ \left|g(z_n)-g(z)\right|<\epsilon_2$$
We define $\varepsilon=\epsilon_1+\epsilon_2$, and therefore
$$\left|f(z_n)-f(z)\right| + \left| g(z_n)-g(z)\right|<\varepsilon$$
If $f(x)=g(x)$ for all $x\in X_\alpha$, then $f(z_n)=g(z_n)$ for all natural $n$, because $z_n\in X_\alpha$. And by triangular inequality we have
$$\left|f(z)-g(z)\right|\leq \left|f(z_n)-f(z)\right| + \left| g(z_n)-g(z)\right|$$
But $\left|f(z)-g(z)\right|$ is absolute value, therefore
$$0\leq \left|f(z)-g(z)\right|<\varepsilon;\ \text{for all $\varepsilon>0$}$$
We claim that $0= \left|f(z)-g(z)\right|$, because if we assume the opposite, we would have $\left|f(z)-g(z)\right|= a >0$, then we could define $\varepsilon=\frac{a}{2}$ and find $a=\left|f(z)-g(z)\right|<\frac{a}{2}$, contradiction.
Hence,
$$0= \left|f(z)-g(z)\right|\implies f(z)=g(z)\ \text{for all $z\in\mathbb{R}$}$$
Ergo $f=g$. $\tag*{$\blacksquare$}$
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What if $X_\alpha$ is not a dense set?
If $X_\alpha$ is not dense set, that is, $\overline{X_\alpha} \subsetneq \mathbb{R}$ is a proper subset. Well, this means that there is some $z\in\mathbb{R}$ such that $z\not\in\overline{X_\alpha}$, or yet, $z$ is not a limit point of $X_\alpha$, then there is no sequence $(z_n)_n$ of $X_\alpha$ such that $z_n\to z$.
 A: Since you included the general-topology tag, I’ll point out that this is a special case of a much more general result that has a short, simple proof.

Theorem. Let $f$ and $g$ be continuous functions from a space $X$ to a Hausdorff space $Y$, and let $D$ be a dense subset of $X$ such that $f\upharpoonright D=g\upharpoonright D$; then $f=g$.
Proof. If not, there is an $x\in X$ such that $f(x)\ne g(x)$. $Y$ is Hausdorff, so there are disjoint open sets $U$ and $V$ in $Y$ such that $f(x)\in U$ and $g(x)\in V$. Let $W=f^{-1}[U]\cap g^{-1}[V]$; clearly $x\in W$, so $W\ne\varnothing$, and the continuity of $f$ and $g$ ensures that $W$ is open in $X$, so $W\cap D\ne\varnothing$. Let $y\in W\cap D$; then by hypothesis $f(y)=g(y)$, but $f(y)\in U$, $g(y)\in V$, and $U\cap V=\varnothing$, so $f(y)\ne g(y)$. This contradiction shows that $f=g$. $\dashv$

It’s not hard to adapt this proof to your specific setting.
A: In fact you can simplify your proof a lot by denoting $h=f-g$. $h$ is continuous as the difference of two continuous maps. You then just have to prove that if a continuous map $h$ vanishes on a dense set $X_\alpha$, then $h$ vanishes on $\mathbb R$. Which is almost immediate. Take $a \in \mathbb R$. As $X_\alpha$ is dense, it exists a sequence $\{a_n\}$ of $X_\alpha$ such that $\lim\limits_{n \to \infty} a_n = a$ and then
$$f(a) = \lim\limits_{n \to \infty} f(a_n) = 0.$$
