False proof that $(c,d)$ is closed. Statement: Prove that $(c,d)$ is a closed set.
Proof:  
Lemma : A point $x$ is a limit point of a set $A$ if and only if $x=\lim_{n\to\infty}a_n $for some sequence $(a_n)$ in $A$ with $a_n\neq x$.
By lemma let $x$ be a limit point of $(x_n)$ contained in given set, then we have $c<x_n<d$ using squeeze principle for limits, $c<x<d$. Thus $x\in A$. Thus the set $A$ has all its limits points in it so its closed.  
The lemma is fool-proof, but I am confused where I went wrong. Please help, for additional information comment below. Thanks a lot.
 A: The squeeze principle for limits, as suggested in comments does not work the way you describe, you can only infer non-strict inequality. Simple counterexample to your claim could be $$-\frac 1 n < 0 < \frac 1 n$$ where if you take limits, you would get $0< 0 < 0$ (which is clearly false).
Actually, the squeeze principle (in its valid form) is equivalent to segments being closed. One direction is obvious, take a convergent sequence $a\leq c_n \leq b$ and apply the squeeze principle to get that $[a,b]$ contains all of its limit points. 
For the other direction, assume that any segment is closed and take convergent sequences $a_n<c_n<b_n$. Denote with $A,B,C$ the appropriate limits. Take any $\varepsilon > 0$. Then, by definition of limit, $c_n\in[a_n,b_n]\subseteq [A-\varepsilon, B+\varepsilon]$ for all but finitely many $n\in\Bbb N$. By closedness of segments, we then have that $C\in[A-\varepsilon, B+\varepsilon]$, and by $\varepsilon$ being arbitrary, we get that $$C\in \bigcap_{n\in\Bbb N}\left[A-\frac 1 n, B+\frac 1 n\right] = [A,B]$$
A: Going into "picky grader mode" and using stars to denote annotations:

By lemma let $x$ be a limit point of $(x_n)$ [*1] contained in given set [*2], then we have $c<x_n<d$ using squeeze principle for limits [*3], $c<x<d$. Thus $x\in A$. [*4] Thus the set $A$ has all its limits points in it [*5, 6] so its closed.



*

*The sequence $(x_{n})$ hasn't been defined (or previously mentioned), so it's awkward to say "$x$ [is] a limit point of $(x_{n})$".


In any case, $x$ should initially denote a limit point of $A$, not of a sequence.


*"given set" presumably means the open interval $(c, d)$ itself.

*The squeeze theorem asserts something like: If $(x_{n})$ is a convergent real sequence with limit $x$, and if $c < x_{n} < d$ for all but finitely many $n$, then $c \leq x \leq d$.
Grammatically, the snippet, "then we have $c<x_n<d$ using squeeze principle for limits, $c<x<d$" appears to assert, "If $x \in (c, d)$ and if $(x_{n}) \to x$, then $c < x_{n} < d$", which is far from true.
If instead this should be parsed as, "then we have $c<x_n<d$. By the squeeze principle for limits, $c<x<d$", the second sentence mis-applies the squeeze theorem, as noted above (and in other comments and answers).

*Item 2. assumed $x \in A$, so this conclusion is tautological.

*This implicitly says every limit point of $A$ is a point of $A$. Instead, one would need to start with an arbitrary limit point $x$ of $A$ and prove that $x \in A$. As several other note, however, this is untrue. (The endpoints of $A$ are limit points of $A$, but not elements of $A$.)

*"[The] set $A$ has all its limit points in it" reads awkwardly as prose. "The set $A$ contains all its limit points" is (to me, at least) smoother. Rough style makes the reader's task more difficult, and the tiring effect accumulates.
Avoiding pronouns and writing mathematical proofs grammatically (including articles, verbs, punctuation) are important habits. Conversely, pronouns and non-grammatical writing in mathematical proofs are generally recipes for ambiguity and misunderstanding. Pronouns, particularly, hide sloppiness. If you can't use a specific noun instead of "it", you've identified a detail you don't fully understand.
A: A set $S$ is closed if $cl(S) = S$, where $cl(S)$ is the closure of $S$, that is, $S$ together with all its limit points. So you must show that every limit point of $S$ is an element of $S$. But $c$ is a limit point of the open interval $(c,d)$ and clearly $c \notin (c,d)$. 
