TITLE: Proving that the set of all increasing sequences is uncountable
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: If you already know that there are uncountably many subsets of $\mathbb N$, then you can solve this problem by exhibiting a one-to-one map from subsets of $\mathbb N$ to increasing sequences of natural numbers, for example the map sending any $X\subseteq\mathbb N$ to the sequence $(x_n)$ where $x_n=2n$ for $n\in X$ and $x_n=2n+1$ for $n\notin X$.