Why does $\mathbb{Z}$ satisfy the ACC? This might be a really trivial question but somehow my reasoning shows the opposite of what it is supposed to be.

Show that the set of integers $\mathbb{Z}$ satisfies the Ascending Chain Condition.

My work:
Since every ideal is of the form $(m)$ for some $m\in\mathbb{Z}$, then as $m$ increases the set $(m)$ becomes less "dense", then how can I show $I_1\subset I_2\subset\dots I_i\subset\dots$? How can a less "dense" set be a superset of a "denser" set?
How can we explain the claim both in rigorous and in intuitive manner?
If there is any flaw in my reasoning, please correct it. 
Thanks!
 A: From your description of your reasoning, I  think you're mixing up $\supset$ and $\subset$, and also perhaps confused about what you want to show.
You're right that as $m$ increases the set $(m)$ becomes less "dense"; in fact, one can be more precise:
$$(m)\subseteq(n)\iff n\mid m$$
where $n\mid m$ means "$n$ divides $m$", or in other words "$n$ is a factor of $m$". Thus, "less dense" sets are contained in "denser" sets, as you would expect. Therefore, given a collection of ideals $I_i=(m_i)$, they are an ascending chain, i.e. they satisfy
$$I_1\subseteq I_2\subseteq I_3\cdots$$
if and only if the integers $m_i$ satisfy
$$m_2\mathsf{\text{ is a factor of }}m_1,\quad m_3\mathsf{\text{ is a factor of }}m_2,\quad\ldots$$
Hopefully it is clear from this why the chain of ideals must stabilize.
You then ask

how can I show $I_1\subset I_2\subset\cdots I_i\subset\cdots$

but this is not the ACC. In fact, the above statement doesn't mean anything - we haven't said what these $I_i$'s are. You want to show that there does not exist such an increasing chain of ideals (or, if you use $\subset$ to mean $\subseteq$, you want to show that any such collection of ideals will eventually stabilize).
A: Think about what containment of ideals in $\mathbb{Z}$ means in terms of prime factorisations. When is $(m) \subseteq (n)$? It means that $n$ has prime factors only occurring in the prime factorisation of $m$, and so as you go "further along" in the sequence you end up either stopping, or finishing at $1$.
It's definitely easier to see this (for me at least) in terms of the equivalent characterisation. The ascending chain condition is equivalent to ideals being finitely generated. In the case of $\mathbb{Z}$, it's even better: they are singly generated.
A: Here is another approach, in its core it is quite the same answer as the others gave, but it's wrapped differently. Its applicability to your situation may depend on theorems which you may or may not have learned about $\Bbb Z$.
Note that if $I$ is a non-zero ideal then $\Bbb Z/I$ is a finite set. If $I\subseteq J$ then $J/I$ is an ideal in $\Bbb Z/I$, and $(\Bbb Z/I)/(J/I)\cong\Bbb Z/J$. Therefore the larger the ideal is -- the smaller the quotient is.
Therefore an increasing sequence of ideals corresponds to a decreasing sequence of finite sets. How long can such sequence be?
A: Hint: $ $ for principal ideals: $\ (a)\supset (b)\!\iff\! a\mid b,\ $ i.e. constains $\!\iff\!$ divides, hence
$$\begin{eqnarray} \ldots && (a_3) &\supsetneq& (a_2)&\supsetneq&(a_1)\quad {\rm wlog}\ \ a_i \ge 0,\ \ {\rm by}\ \ (a) = (|a|) \\
\iff\ \ldots && \ \ a_3 &\,\mid& \ \ a_2 &\,\mid& \ \ a_1\quad\ \ \text{where each division is  proper}\\
\Rightarrow\,\ \ \ldots && \ \ a_3 &<& \ \ a_2 &<& \ \ a_1 \\
\end{eqnarray}$$
Thus the chain stabilizes at the ideal with the least generator $\ge 0,\,$ which is also the least element in the (ascending) union of the ideals. Viewed arithmetically, this least element is the gcd of the generators, and also the gcd of all elements in the union of the ideals. 
A: How can a less "dense" set be a superset of a "denser" set?
Thinking of chain conditions as "density" does not really strike me as a good mental model. It really should be something more like "size."
Chain conditions like the ACC and DCC just put limits on "how deep" or "how tall" a chain can be. In $\Bbb Z$, you are trying to show that you can't have an infinitely tall chain of proper containments. That is, it's a limit on how fast such chains can grow in a particular direction.
Why is this? Since $\Bbb Z$ is a PID, its ideals are all principal and have particularly simple relationships with each other: $(b)\subseteq (a)$ iff $a$ divides $b$. But the fundamental theorem of arithemetic puts limits on what can divide $a$, and hence limits what can properly lie above $(b)$.
But on the other hand, $\Bbb Z$ admits infinitely deep descending chains. For example $(2)\supsetneq(4)\supsetneq(8)\supsetneq\dots$ is possible. If it were the case that $(2^n)=(2^{n+1})$, then $2^{n+1}|2^n$ is just not possible because of the nature of the integers.

A good mental model of density is something like the field of fractions for a domain. For example $\Bbb {Z}$ is really "dense" (="large") in its field of fractions $\Bbb Q$ since every nonzero additive subgroup of $\Bbb Q$ has to have nonzero intersection $\Bbb Z$. In that sense $\Bbb Z$ is so "thick" in $\Bbb Q$ that you can't avoid hitting it.

To get a better flavor of the ACC and DCC, I recommend you search the site for some examples of an Artinian module which isn't Noetherian, so you can see what that example looks like. If you have access, you can also skim Lectures on Modules and Rings by Lam, page 228, which is a section on finiteness conditions on rings. There you can find many interesting variants of chain conditions, which will hopefully convince you that such conditions aren't really about density, but rather about height and depth.
A: Suppose we have a sequence of increasing subgroups $G_n = (k_n)$ that doesn't stabilize. Then $G_i = (k_i) \subset G_j = (k_j)$ iff $k_j|k_i$. Then all $k_j$ divide $k_1$, but we cannot have infinitely many distinct $k_j$ dividing $k_1$.
