Proof of sequence limit, using epsilon-delta method. ${\displaystyle (a_n)}$ is a sequence with ${\displaystyle a_n = \frac{1}{\sqrt{n}}}$. Proove that ${ \displaystyle \lim\limits_{n\to\infty}{a_n} = 0 }$, using epsilon-delta method.

First of all, I assume that ${ \displaystyle n \in [1;+\infty) }$.
For all ${n}$, which are larger or equal to ${1}$, ${n^{th}}$ term of sequence is strictly larger than zero and less or equal to one.
${ \forall n \geq 1 : 0 < a_n \leq 1  }$
Assume some ${ \displaystyle \epsilon > 0 }$, such that ${ \displaystyle 1 - \frac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow \left(\frac{1}{1-\epsilon}\right)^2 > n }$
Now let ${ \displaystyle N = 1 + \left\lceil \left(\frac{1}{1-\epsilon}\right)^2 \right\rceil }$, then ${ \displaystyle \left| \frac{1}{\sqrt{n}} - 1 \right| < \epsilon }$ ${ \displaystyle \forall n \geq N }$

Have I understood the task? Is my solution correct?
 A: 
Assume some $ϵ>0$, such that $1−\frac1{\sqrt n}<ϵ$ […]

This hints at a serious misunderstanding. You appear to be assuming what it is that you have to prove. 
"The enemy" comes with a devilishly clever choice of $\epsilon$ and you have to show that you can find an $N$ that demonstrates that "the enemy's" clever choice isn't good enough. 
A: As mentioned in comments, you should be investigating the inequality $\frac{1}{\sqrt{n}}<\epsilon$, not $1-\frac{1}{\sqrt{n}}<\epsilon$. 
It's OK to do preliminary side work that establishes $$\frac{1}{\sqrt{n}}<\epsilon\iff n>\frac{1}{\epsilon^2}$$ and then use that to decide that a good choice for $N$ is $\lceil\frac{1}{\epsilon^2}\rceil$. But the logic of an epsilon-N argument doesn't start until:


*

*Let $\epsilon$ be someone's arbitrary idea of what "small" is. That is, let $\epsilon$ be some positive number that is handed to me by someone else.

*Turn around back to them and say "based on your $\epsilon$, here is an $N$-value:  $\lceil\frac{1}{\epsilon^2}\rceil$. And I can demonstrate that '$n\ge N\implies \frac{1}{\sqrt{n}}<\epsilon$'."


That last directional implication must be in that direction for the logic of this argument to work.
A: There are a few errors in what you have written. First you want to prove that the limit is $0$. In which case, you want to prove the following.
Given any $\epsilon >0$, there exists a $N(\epsilon) \in \mathbb{N}$,
$$\text{ such that for all $n > N(\epsilon)$, we have }\left \vert a_n - 0 \right \vert < \epsilon \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$
When you are proving a limit from first principles, you cannot choose $\epsilon$; $\epsilon$ is given to you. The challenge for you is to find $N(\epsilon)$. So whatever be the '$\epsilon$' I give you, you should be able to find a $N(\epsilon)$ such that $(\star)$ is satisfied.
In your problem, given $\epsilon > 0$, you need to find $N(\epsilon) \in \mathbb{N}$,
$$\text{ such that for all $n > N(\epsilon)$, we have }\left \vert \dfrac1{\sqrt{n}} - 0 \right \vert < \epsilon \text{ i.e. } \dfrac1{\sqrt{n}} < \epsilon $$
Now note that $a_n = \dfrac1{\sqrt{n}}$ is a decreasing sequence and hence if we find $m$ satisfying $(\star)$, for any $n > m$, $(\star)$ will be satisfied since for $n>m$, we have $a_n < a_m$. This motivates us to choose $N(\epsilon)$ as $$N(\epsilon) = \left \lceil\dfrac1{\epsilon^2} \right \rceil$$ Note that $$N(\epsilon) = \left \lceil\dfrac1{\epsilon^2} \right \rceil \geq \dfrac1{\epsilon^2} \implies \sqrt{N} \geq \dfrac1{\epsilon} \implies \dfrac1{\sqrt{N}} \leq \epsilon$$
Hence, for all $n > N$, we have that
$$\dfrac1{\sqrt{n}} < \dfrac1{\sqrt{N}} \leq \epsilon$$
