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: First, we show that $f(E)$ is dense in $f(X)$. Note that it might be that some members of $f(X)$ also belong to $f(E)$ which means that we only need to show that those members of $f(X)$ that do not additionally belong to $f(E)$ are limits points of $f(E)$. To this end, let $y \in f(X)$ such that $y \notin f(E)$. Then, $y = f(p)$ for some $p \in X \setminus E$. Since $E$ is dense in $X$, $p$ is a limit point of $E$. Since $f$ is continuous at $p$, for any $x \in E$ satisfying $x \to p$, we have that $f(x) \to f(p) = y$. Since $y$ was arbitrarily chosen, $f(E)$ is dense in $f(X)$.
My question: Is my proof for showing that "$f(E)$ is dense in $f(X)$" inaccurate in any way?
It seems fine to me, and assuming this is your first time learning analysis, I would try to be more explicit by saying this for the last part: since $p$ is a limit point of $E$, there exists a sequence of points $\{x_n\}\subset E$ with $x_n\rightarrow p$. Then by continuity we get a sequence of points $\{f(x_n)\}\subset f(E)$ with $f(x_n)\rightarrow f(p)=:y$. Hence every neighborhood of $y\in f(X)$ contains a point of $f(E)$ and $f(E)$ is dense in $f(X)$. $\boxed{}$