Let $f:X\rightarrow Y$ be a continuous function,and let $E$ be a dense subset of $X$.
Prove that $f(E)$ is dense in $f(X)$.
I know the solution, but a few details are unclear to me:
We proceed by proving that every open subset of $f(X)$ intersects $f(E)$. Let $V$ be an open subset of $f(X)$. There exists $V'$ open in Y, such that $V' \cap f(X) =V$ (How do we know that there exists an open V' in Y, and how do we know that it satisfies $\boldsymbol{V' \cap f(X) =V}$?). Since $f$ is continuous, $f^{-1}(V')$ is open in $X$. Then $f^{-1}(V') \cap E \neq \emptyset$ because $E$ is dense in $X$. This implies that $f(f^{-1}(V')) \cap f(E) \neq \emptyset$. But $f(f^{-1}(V'))\subset V'$, and so $\boldsymbol{V \cap f(E) =V' \cap f(E)}\neq \emptyset$; QED. (How did we arrive at the bolded inequality?)
The first bolded statement is just the definition of subspace topology: if you have a topological space $A$ and a subset $B \subset A$ then according to the said topology a subset $V \subset B$ is open if and only if there is an open subset $U$ of $A$ such that $V= B \cap U$. Thus being $V \subset f(X)$ open there must be an open subset $V'$ of $Y$ such that $V= V' \cap f(X)$.
The bolded equality at the end is just a basic conclusion of set inclusions: being $f(E)$ a subset of $f(X)$ we get $V' \cap f(E)= V' \cap f(E) \cap f(X)$, but now $V' \cap f(X)= V$ and we obtain $V' \cap f(E)= V' \cap f(X) \cap f(E)=V \cap f(E)$.