Equivalence of definitions of the closure of a set I have recently stumbled across a proof in a book about functional analysis concerning closure of the set and I can't seem to grasp one part of it.
Let $X$ be a normed space over $\mathbb{K}$ and $M$ be a subset of $X$. Then, we define the closure $\overline{M}$ of $M$ as
$$
\overline{M}:= \text{smallest closed set of $X$ containing $M$}.
$$
Then, the proposition states that for $M$ being a nonempty subset of $X$, the following holds:
$(i)$ $u\in \overline{M}$ iff, for some sequence $\{ {u_n}\}$ in $M$
$$
u_n \to u \quad \text{as} \quad n\to \infty.
$$
In the proof the author denotes $C$ as the set of all points $u\in X$ such that $u_n\to u$ as $n \to \infty$ for some sequence $\{ {u_n} \}$ in $M$. Then, he shows that $C$ is closed and that part of the proof is clear to me. However, for the converse implication he proceeds as follows:
Let $\mathcal{C}$ be a closed subset of $X$ such that $M\subseteq \mathcal{C} $. Then
$$
C \subseteq \mathcal{C}.
$$
Therefore, $C$ is the smallest closed subset that contains the set $M$, i.e., $C=\overline{M}$.
I can't see where does the statement $C\subseteq \mathcal{C}$ come from and would be very thankful for any explanation of it.
 A: Take $x \in X\backslash\mathcal{C}$. There exists an open ball $B \ni x$ such that $B\cap M = \emptyset$. Therefore $x\in X\backslash C$ and we have shown that $C \subset \mathcal{C}$.
A: Let $(X, \|\cdot\|)$ be the normed space. You want to show that $C \subseteq \mathcal C$. 
Let $u \in C$. Then you know there exists a sequence $\{u_n\}$ in $M$ such that $u_n \to u$. 
Suppose for the sake of contradiction that $u \not \in \mathcal C$, that is $u \in X \setminus \mathcal C$. But $X \setminus \mathcal C$ is open (since it's the complement of a closed set) and hence there exists $\varepsilon > 0$ such that $B_\varepsilon \subsetneq X \setminus \mathcal C$. For this $\varepsilon > 0$ there exists $N_\varepsilon \in \mathbb{N}$ such that for all $n \geq N_\varepsilon$ you have $\|x_n - x\| < \varepsilon$, that is $x_n \in B_\varepsilon(x)$. But since $M \subseteq \mathcal C$ you have $M \cap B_\varepsilon(x) = \emptyset$ and this contradicts the assumption $x_n \in M$ for all $n \in \mathbb{N}$.
