Subset is closed IFF limit of every sequence $x_{n}$ results in x in the subset 
Proposition:
A subset $U \subseteq  X$ is closed IFF for every sequence $x_{n}\rightarrow x$ with $x_{n} \in U$
for all n we also have $x \in U$.

The way in which the proof started out puzzles me. I've never seen proof like this before so either there is a mistake in the proposition from which my notes came from or I am misunderstanding the proposition.
Proof: First suppose U is closed and $x_{n}\rightarrow x$ with $x_{n} \in U$.
Right off the bat, the proof baffles me. From my knowledge, it is perfectly fine to assume first that $U \subseteq X$ is closed and then show that it leads to the fact that for every sequence $x_{n}\rightarrow x$ with $x_{n} \in U$
for all n, we also have $x \in U$.
But what the proof seems to be now is that the author assumes both the antecedent and the consequences.
A bit of clarification would be helpful.
Thanks in advance.
 A: Theorem:
$\forall U\subseteq X,~\left(U\text{ is closed }\displaystyle\Longleftrightarrow\forall \{x_n\}\subseteq U,~(\lim_{n\to\infty}x_n=x\Rightarrow x\in U)\right)$
Proof:
Step 1: Since this is a universal quantification. The regular way to prove such universal quantification is first let (any) $U\subseteq X$, and show: 

$U$ is closed$\displaystyle\Longleftrightarrow\forall \{x_n\}\subseteq U,~(\lim_{n\to\infty}x_n=x\Rightarrow x\in U)$

(if this holds, then we know "$\forall U\subseteq X,\cdots$(the yellow part)" holds.)
Anyway, we next want to show (WTS) the yellow part.
Step 2: We first prove "$\Rightarrow$" holds(it's exactly what you ask, so the other direction is omitted):
Suppose $U$ is closed, WTS 

$\displaystyle\forall \{x_n\}\subseteq U,~(\lim_{n\to\infty}x_n=x\Rightarrow x\in U)$

Step 3: Next, the universal quantification again! Let $\{x_n\}\subseteq U$, WTS

$\displaystyle\lim_{n\to\infty}x_n=x\Rightarrow x\in U$

Step 4: Next again, we suppose $\displaystyle\lim_{n\to\infty}x_n=x$, and WTS

$x\in U$

Final step:
if you have shown  $x\in U$, then it's all done.
