I'm trying to prove:
$f: X \to Y$ is continuous, $g: X \to Y$ is continuous, $E \subset X$ is dense in $X \implies$ $f(E)$ is dense in $f(X)$ and $\forall p \in E, g(p) = f(p) \implies g(p) = f(p) \forall p \in X$.
My attempt for proving $f(E)$ is dense in $f(X)$:
Suppose $f: X \to Y$ is continuous, $g: X \to Y$ is continuous, $E \subset X$ is dense in $X$. Let $f(x) \in f(X)$ for some $x \in X$. Let $f[V_a(x)]$ be an arbitrary neighborhood of $f(x)$.
Since $E$ is dense in $X$, $x$ must be a limit point of $E$, that is so that, for the neighborhood $V_a(x)$ of $x$, $\exists q \ne x$ such that $q \in E$ and $q \in V_a(x)$. Then, $f(q) \ne f(x)$ is such that $f(q) \in f(E)$ and $f(q) \in f[V_a(x)]$ which shows that $f(x)$ is a limit point of $f(E)$. Since $f(x)$ was arbitrarily chosen, $f(E)$ is dense in $f(X)$.
My attempt for proving $\forall p \in E, g(p) = f(p) \implies g(p) = f(p) \forall p \in X$:
We prove the contrapositive: suppose $g(p) \ne f(p) \forall p \in X$. Since $E \subset X$, $g(p) \ne f(p) \forall p \in E$ follows immediately.
My question: I am sure there is plenty wrong with my proofs (for instance, I never used the continuity of $f$ and $g$). I want to ask how my proofs are incorrect. Please know that I'm not looking for new ways of proving this statement as those can be found on the internet. I'm looking for feedback on my proofs, and ways of improving them. Thanks!