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By definition, the function $f$ from $X$ to $Y$ is a bijection if every element $y$ in $Y$ is a value of $f(x)$ for exactly one $x$ in $X$. Also by definition, if there is a bijective correspondence between the sets $S$ and $\mathbb{N_m}$ then S has cardinality $m$, which can be denoted by $|S|=m$.

I want to prove that if we have a set S such that $|S|=s$ and $|S|=t$, then $s=t$.

Given is that there are bijections $f:S\to\mathbb{N_s}$ and $g:S\to\mathbb{N_t}$. Now we can use the pigeonhole principle for the composite functions $f\circ g^{-1}$ and $f^{-1}\circ g$ to obtain $s\le t$ and $t\ge s$, respectively.

Now my question is: why can we not say directly, from the definition of a bijection, that the cardinality of $X$ and $Y$ are equal. So that we can prove it by saying that if $|S|=s$ and $|S|=t$ then $s=t$ because both $f$ and $g$ are bijections with the same domain.

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    When you say "directly", there is actually quite a lot going on. The proposed solution tries to reduce the amount of implicit argumentation. – 2017-01-24
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    See Schroder-Bernstein Theorem – 2017-01-24
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    What definition of cardinality do you use? – 2017-01-24
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    Basically, we define the reation of *equal cardinaloty* betweenntwo sets $X$ and $Y$ by way of the bijection $f : X \to Y$. Then we say that a *finite* set $X$ has $n$ elements if there is a bijection $f : X \to I_n$, where $I_n = \{ 1,2,\dots n \}$. – 2017-01-24
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    So the issue is : is it possible that $f : X \to I_n$ and $g : X \to I_m$, when $n \ne m$ ? – 2017-01-24
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    @skyking The definition of cardinality is in the second sentence, where $\mathbb{N_m} = {1,2,...,m}$. – 2017-01-24
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    Usually, in math the answer : "Obviously not" is not liked. – 2017-01-24
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    @Arthur I was told that the proposed solution was a circular argument. Do you have any idea why that is so? – 2017-01-24
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    You have to note that $|S|=m$ and $|S|=n$ means that we have bijections : $f: S \to I_m$ and $g : S \to I_n$. Thus, we can build a bijection $b : I_m \to I_n$, but if we rely on this fact to assert that $m=n$, we are in a circle, because this is exactly what we want to prove. – 2017-01-24
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    You have to consider that the theory of *cardinality* is general: it is the same for finite and infinite sets. If we start assuming that in order to know "how many" elements there are in a set it is enough to count them (and this was enough for quite a long time) we do not need a specific theory. But then we will be in trouble with infinite sets, because we have to conclude (by an obvious observation) that the number of even numbers is half the number of numbers... – 2017-01-24
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    @MauroALLEGRANZA Thank you very much for your replies. But you say that we can build a bijection $b : I_m\to I_n$. Does it not help that $|I_m|=n$, and similarly if we exchange $m$ and $n$ that $|I_n|=m$? – 2017-01-24
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    @MauroALLEGRANZA Perhaps, perhaps not. Sometimes it's educational to walk the same evolution as math (which means start with natural numbers and then onto infinite cardinals). Note that one aproach in defining the concept is to avoid defining the actual cardinal number, but just define the corresponding operations and relations. That way the question would be rather to show that equicardinality is a equivalence relation (in the broader sense). – 2017-01-24
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    Assuming that we agree on the "pedagogical benefits" of proving this "obvious fact", the issue is: from $b : I_m \to I_n$ we have $|I_m|=n$ and from $b' : I_n \to I_m$ we have $|I_m|=n$. But how we conclude that $m=n$ ? We can do so because we "know" that the number of elements of a set is univocally defined... and this is exactly waht we are asked to prove. So, again, the rule of the game is "prove it" (in a non-circular way). – 2017-01-24
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    @MauroALLEGRANZA Okay, I think I am starting to grasp it now. Thank you. – 2017-01-24

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What you want to prove is that if $\mathbb N_s\cong\mathbb N_t$ then $s=t$. WLOG we can assume that $s\le t$.

Basically you may prove this by showing that there can be no smallest such cardinality that allows this, because if you find one you can create a case with one element less - unless one is $\emptyset$. You end up with the claim that $\emptyset\cong X$ for some non-empty $X$.

Formally one starts with showing that cardinality is a total order. So if $X\prec Y$ and $Y\preceq Z$ we have that $X\prec Z$. We can use this to show that if $\mathbb N_s \prec \mathbb N_{s+1}$ then it's true for all $t>s$ that $\mathbb N_s\prec\mathbb N_t$. That is we only have to show that $\mathbb N_s \prec\mathbb N_{s+1}$.

Next is to use induction. Obviously $\emptyset=\mathbb N_0\prec\mathbb N_1$ (since to have a bijection with the empty set the other need to be empty). Now assume it's true for some $s$. Now assume that we have $\mathbb N_{s+1}\cong\mathbb N_{s+2}$ that is there's a bijettion $\Phi$, but then we can restrict that to $\phi:\mathbb N_s\to\Phi(\mathbb N_s) = \mathbb N_{s+2}\setminus \{\Phi(s+1)\}$. Now we use this latest form to form a bijection to $\mathbb N_{s+1}$ which would show that $\mathbb N_s\cong\mathbb N_{s+1}$ which we assumed not to be possible.