Claim: Prove that the set of all increasing sequences $n\_1<n\_2<...<n\_n<...$ is uncountable. 

Here's my attempt:

Let $X=\{(x\_n)\_{n\in N};(x\_n)\_{n\in N}$ is increasing$\}$. Consider $f:N\rightarrow X$. We claim that no $f$ can be surjective.

Denote $(u\_n)\_{n\in N}$ the image of $f$ at the point $u\in N$, ie, $f(u) =(u\_n)\_{n\in N}$ . We construct a sequence $(s\_n)\_{n\in N}$ such that $(s\_n)\_{n\in N}$$\neq f(s)$ for all $s\in N$. 

For, construct $(s\_n)\_{n\in N}$ taking $s\_1=\prod\_{k=1}^{\infty} f(k)\_1$, $s\_2=\prod\_{k=1}^{\infty} f(k)\_2$, ... , $s\_n = \prod\_{k=1}^{\infty} f(k)\_n$, ...

By the monoticity of multiplication, we guarantee that $(s\_n)\_{n\in N} \in X$. But by the way we built it, it is such that $(s\_n)\_{n\in N} \neq f(s)$ for all s natural. Hence, $f$ cannot be surjective and therefore X is not countable.

Is it correct? And if yes, i find pretty intuitive that $(s\_n)\_{n\in N} \neq f(s)$ for all s natural. But can someone turn it out more rigorous and clear?

EDIT:

I came up with a new solution and better notation. Here it is:

Let $X=\{(x\_n)\_{n\in N};(x\_n)\_{n\in N}$ is increasing$\}$. Consider $f:N\rightarrow X$. 

Denote $(u\_n)\_{n\in N}$ the image of $f$ at the point $u\in N$, ie, $f(u) =(u\_n)\_{n\in N}$ . We may write this sequence as $f(u)= (f(u)\_1,...,f(u)\_n,...)$. Also define $A\_u=\{f(u)\_1,...,f(u)\_n,...\}$ as the set of all elements of the sequence $f(u)$. We define this set for every $u \in N$.

Now we construct a new sequence $(s\_n)\_{n\in N}$ inductively given by:

Define $f(s)\_1 $= the smallest element of $A\_1$.

Supposing defined $f(s)\_1,...,f(s)\_n$, let $S\_{n+1} = \{x \in A\_{n+1}; x>y$, where $y$ is the smallest element of $A\_{n}\}$.

Then take $f(s)\_{n+1}$= smallest element of $S\_{n+1}$.

Therefore the sequence $(s\_n)\_{n\in N}$ is now defined inductively, and by construction $(s\_n)\_{n\in N}$ is increasing. Therefore, $(s\_n)\_{n\in N} \in X$ and $(s\_n)\_{n\in N}$ is such that $(s\_n)\_{n\in N} \neq f(s)$, for all $s \in N$. Therefore $f$ is not surjective and hence X is not countable.