$\left(X_n\right) \to x > 2$. Prove that $\,X_n > 2$ for all $n\ge K$ $\left\lbrace X_n\right\rbrace$ is a convergent sequence whose limit is $x > 2$.
How do I prove that there exists $K\in \mathbb N$ such that $X_n > 2$ for all $n\ge K$?
I have learnt the $\varepsilon-K$ definition of limits, and applying to this question:
$$\varepsilon>0,\; \exists\, K\in \mathbb N\quad\text{ such that }\quad \left\lvert X_n - x\right\rvert < \varepsilon\;\;\forall n \ge K.$$
I can only think of removing the modulus and then adding $x$ to all sides to get: $x-\varepsilon < X_n < x+\varepsilon$.
How do I proceed?
 A: If $X_n$ is convergent, then its limit exists. Let $x$ be such a limit. Since $x>2$, the distance between $x$ and $2$ is $D=x-2>0$. 
By the definition you wrote of the limit of a sequence, you can pick any $\epsilon$, which you can think as the distance between $X_n$ and $x$ you desire, and you'll get an $N$ such that for all $n>N$, the distance $X_n$ is closer to $x$ than $\epsilon$. 
In this case, you'll want to pick $\epsilon=D$ (or any positive number less than $D$). By the definition of the limit, you'll get an $N \in \mathbb{N}$ that ensures what you wanted.
To see why intuitively, I think you should draw a picture of the real line, with a dot on the $2$, another one in $x$, the distance $D$, and also the meaning of the limit (as $n \rightarrow \infty$, then $X_n \rightarrow x$).
A: By definition of limit of sequence
\begin{align}
x &= \lim_{n\to\infty} X_n &\iff&& \forall \,\varepsilon>0 \; \exists\, K\in\mathbb N \,: \;\; \left\lvert  X_n - x \right\rvert &< \varepsilon\;\forall \,n>K
\end{align}
Since limit $x>2$ we know that $x-2>0$, so we can use it as $\varepsilon$ in the definition of limit: 
\begin{align}
\varepsilon &:= x-2>0 & \implies && \exists\, K_0\in\mathbb N \,: \;\; \left\lvert  X_n - x \right\rvert&< \varepsilon= x-2\;\forall \,n>K_0
\end{align}
We can rewrite inequality $\,\left\lvert  X_n - x \right\rvert < \varepsilon\,$ as following:
\begin{alignat}{4}
 \left\lvert  X_n - x \right\rvert &< x-2 \quad 
 & \iff &\quad& - (x - 2) &< X_n - x &&< x-2
\\ && \iff && 2 &< X_n &&< 2x-2
\end{alignat}
Thus we conclude that there exists natural number $K$ such that $X_n>2$ whenever $n>K$.
We can rewrite this conclusion in a more formalistic way:
$$
\bbox[1ex,border:solid 2pt #e10000]{\lim_{n\to\infty} X_n = x>2\hspace{3ex}  \implies \hspace{3ex} \exists\, K\in \mathbb N : \hspace{2ex} \forall \,n>K \hspace{2ex} X_n > 2\phantom{\sum^{\infty}}\hspace{-2ex}}
$$
A: Given $\varepsilon >0$, there exists $N>0$ such that $ X_n=x>2$, whenever $n>N$. Thus, we have $X_n>2$, whenever $n>N$.
Comment
Here, I am using the fact that $a=b \iff \forall \varepsilon >0   \ \   |a-b|<\varepsilon$
Take this as shortest answer to your question. Thanks.
