There exists infinitely many irrationals within a range of reals? Theorem: There exists infinitely many irrationals in a range of reals.
Proof:
Let the range be $R=(a,b)$ where $a<b$ and $a,b\in\mathbb{R}$
Suppose some numbers $n_1\ldots n_k\in R$ where each number is greater than the one before, then $a<n_1<\cdots<n_k<b$. From this it follows
\begin{align}
a&<n_1\\
n_1&<n_2\\
&\vdots\\
n_{k-1}&<n_k\\
n_k&<b
\end{align}
So
$$
a+\sum\limits_{i=1}^k n_i<b+\sum\limits_{i=1}^k n_i\qquad\therefore a<b
$$
Because $k$ can be arbitrarily big, and the truth values of $a<b$ does not depend on $k$, we can conclude that there are infinitely many numbers on a given range. We will now use the fact that the distance between two numbers -irrationals too- is always finite.
Because there are infinitely many numbers in a range where $a<b$ and infinity can be divided into infinite finite parts, then there must be infinite irrationals in that range. $\blacksquare$

My question is, is this argument valid? Any not-so-right steps?
 A: No, your proof misses the point.  Aside from the fact that you start with an assumption about the existence of a set of ordered intermediate values of arbitrary size, and spend most of your time proving that $a<b$, which was already given, the problem is that what you have shown is that there are infinitely many reals between any two unequal reals.  But how do you know that almost all of them aren't rational, leaving only finitely many irrationals?
I would start by proving that between any two rational numbers $q_1$ and $q_2$  with $q_1<q_2$ there exists at least one irrational, by proving that 
$q_1+\frac{\sqrt{2}}{2}(q_2-q_1)$ is irrational.  (The proof of that is a lot like the
proof that $\sqrt{2}$ is irrational.)
Then I would use the fact that at least one rational number must exist between any two unequal reals (proved by using a denominator bigger than twice the reciprocal of the distance between the reals).  Apply that to $a$ and $b$, getting a rational $q_x$, and apply it again between $a$ and $q_x$ getting another rational $q_y$.  
Now for any given $N$ you can break up the interval between $q_x$ and $q_y$ into $N+2$ rational sub-intervals and by the first part of the proof you can slot $N+1>N$ irrationals, one in each interval.  So for any arbitrary $N$ there are more than $N$ irrationals in the interval.
