I want to ask if this demonstration is correct.
I want to prove if we have two topological spaces $(X,\tau)$ and $(Y,\tau')$ ($(Y,\tau')$ is Hausdorff) and we have two continuous functions $\phi$ and $\phi'$ that are equal in a dense subset Z of X then the two functions are equal in all X.
I prove this:
We suppose that there exists c $\in X$ such that $\phi(c) \neq \phi'(c)$. Since $(Y,\tau')$ is Hausdorff, there exist $G,G' \in (Y,\tau')$ such that $\phi(c) \in G, \phi'(c) \in G'$ and $ G \cap G' = \varnothing.$ But since $\phi$ and $\phi'$ are continuous, $\exists U,U' \in (X,\tau)$ that satisfy c $\in U, c \in U'$ and $\phi(U) \subseteq G, \phi'(U') \subseteq G'$.
The subset $U\cap U' \neq \varnothing $ (because $c \in U , U'$) and $(U\cap U')\cap Z \neq \varnothing$ then $\exists z\in U,U',Z$ but then $\phi(z) = \phi'(z)$ and $G\cap G' \neq \varnothing$, and this is a contradiction because we have that $G\cap G' = \varnothing$.
The proof is perfectly fine and direct.
An alternative proof: $Y$ is Hausdorff iff $\Delta_Y =\{(y,y): y \in Y\} \subseteq Y \times Y$ is closed.
$\phi, \phi'$ being continuous means that $\phi \nabla \phi' : X \to Y \times Y$ defined by $(\phi \nabla \phi')(x) = (\phi(x), \phi'(x))$ is continuous.
Then $\phi = \phi'$ on $Z$ can be expressed as $Z \subseteq (\phi \nabla \phi')^{-1}[\Delta_Y]$ and as the right hand side is closed and $Z$ is dense, $(\phi \nabla \phi')^{-1}[\Delta_Y] = X$, and $\phi = \phi'$ on all of $X$.