Showing that the closure of a subset of a closed set is going to remain a subset of the closed set? It is clear intuitively that the closure of a subset of a closed set will remain a subset of the original closed set. However, I am struggling to prove that this is the case.
For example, in a problem I have been working on, I have been trying to show that $[0,1] \subset $ cl$(A)$ for $A := \mathbb{Q} \cap [0,1]$
We know that cl$( \mathbb{Q} \cap [0,1]) = \bigcap\limits_{\delta > 0} A_{\delta} \space$ for $A_{\delta} = \bigcup\limits_{x \in A} B(x, \delta)$ where $B(a,r)$ is a ball of radius $r$ centered at $a$.
This reduces the problem down to one of set manipulations, but I am struggling to construct a proof that inclusion holds (despite having a good intuition for it).
I have managed to show that the converse inclusion holds, however, I am trying to show equality here, and so I need to show this other inclusion direction to complete the argument.
I would be grateful for any assistance or references that could help here. To clarify, I would like to show this using the definition of the closure specified above.
 A: The closure of a set is the intersection of all closed sets containing it, so the closure of a set is a subset of any closed set containing it.
Using the suggestion of @Aaron, and your definition of $cl(A)$ in some metric topology:
Part 2)
\begin{align*}
A \text{ closed} &\implies \forall y \notin A \quad \exists \delta>0 \quad B(y,\delta) \cap A = \emptyset \\
&\implies \forall y \notin A \quad \exists \delta>0 \quad \forall x \in A \quad B(x,\delta/2) \cap B(y,\delta/2) = \emptyset \\
&\implies \forall y \notin A \quad \exists \delta>0 \quad y \notin A_{\delta/2} \\
&\implies \forall y \notin A \quad y \notin cl(A) \\
&\implies cl(A) \subset A\\
\end{align*}
Since we always have $A \subset cl(A)$, which follows easily from the definition of $cl(A)$ that we're using, this gives us: $A \text{ closed} \implies cl(A)=A$.
Part 3)
$A \subset B \implies \forall \delta > 0 \quad A_\delta \subset B_\delta \implies cl(A) \subset cl(B)$
Using these two parts, we get that if $A$ is a subset of a closed set $B$, then $cl(A) \subset cl(B) = B$
