Prove that if $X \subset F$ and $F$ is closed then $\bar{X} \subset F$.
Remark: This is from a first course in real analysis so all sets are implicitly assumed to be subsets of $\mathbb{R}$ and all proofs should be rather elementary. Also I don't know if this is standard notation but $\bar{X}$ is the notation my textbook uses for the closure of $X$.
Attempt:
$X \subset F \Rightarrow$ all elements of $X$ are in $F$.
$F$ is closed $\iff F = \bar{F} \iff$ all adherent points of $F$ are in $F$.
Suppose that there exist elements of $\bar{X}$ that are not in $F$. Then, there exist points which are adherent to $X$ but are not adherent to $F$, which contradicts the hypothesis that $F$ is closed. Therefore we must have that $\bar{X} \subset F$.
Best Answer
Suppose $X \subset F$. Then, if $x$ is adherent to $X$, every open set containing $x$ intercepts $X$ at some point, so this implies that every open set containing $x$ intercept $F$ at some point, and we conclude $\bar{X}\subset\bar{F}$. Thus, we have $\bar{X}\subset \bar{F} = F$.