I think you may be confusing a few things here.
First you speak of the set of all functions $\mathbb N \to \{0,1,2\}$, in other words sequences with values in $\mathbb N$, since such a function can be seen as the sequence $ f(0) , f(1) , f(2), f(3),\dots $ (I'm assuming $0\in\mathbb N$, unaware of the convention you may use.)
For instance, one such function may be $0,1,0,1,0,1,...$ where $f(n)$ is $0$ if $n$ is even and $f(n)=1$ if $n$ is odd. Let's call the set of all these functions $S$. Indeed $S$ is uncountable as can be seen with the standard diagonal argument: if we had a list $f_1,f_2,f_3,...$ containing each and every of these functions /sequences once, we could create a new function/sequence not on the list, by making it differ from $f_1$ in the first position, from $f_2$ in the second and so on.
The other thing you are talking about are functions that map $1$ to $1$ and all others to $2$. This means the only choice you have left when creating such a function is it's value at $0$, which can be $0$, $1$ or $2$. So we find $3$ such functions: $ 0,1,2,2,2,2,\dots $ $ 1,1,2,2,2,2,\dots $ $ 2,1,2,2,2,2,\dots $ So the cardinality of the set $T$ equals $3$, which is a finite number. $T$ fits in $S$ nicely, so there's no problem here.
Either way, it's no contradiction to have a countable subset in an uncountable set. For instance, the natural numbers are a countable subset of the real numbers.
-- edit: In fact, one can show that every infinite set, has a countable subset. On the other hand, a countable set cannot have an uncountable subset.