Clarification wanted: Let $T$ be the set of all infinite sequences of $0$'s and $1$'s with finitely many $1$'s. Prove that $T$ is denumerable. I think I'm misunderstanding the following proposition.

Let $T$ be the set of all infinite sequence of $0$'s and $1$'s with only finitely many $1$'s. Prove that $T$ is denumerable. 

I'm also given the following lemma to use:

Let $\{A_{i}| i\in\mathbb{N}\}$ be a denumerable family of finite non-nonempty sets which are pairwise disjoint, then $\bigcup _{i\in\mathbb{N}}A_{i}$ is denumberable. 


I'm not quite seeing how adding the condition that there are finitely many $1$'s makes this the set $T$ countable, since there is still a possibility of infintely many $0$'s. It still looks susceptible to the diagonalization argument. 
Could anyone clear up why the the set is countable? 
 A: The other answers address the issue of showing that the set is countable; let me try to tackle the other end, your intuition that the set should be uncountable. You write:

"It still looks susceptible to the diagonal argument."

OK, let's see if it actually is. The diagonal argument is quite non-intuitive at first; if you think it applies, it's always good to write it out in detail and see if it actually works.

So fix an enumeration $s_n$ of infinite binary sequences with only finitely many $1$s. We define the diagonal sequence $D$ as $$D(n)=1-s_n(n).$$
Now, certainly $D\not=s_n$ for any $n$. So, in order to show that $\{s_n: n\in\mathbb{N}\}$ isn't the set of all infinite binary sequences with finitely many $1$s, we just need to show:

$D$ is an infinite binary sequence with only finitely many $1$s.

But is this true? Why or why not? 
And then: if $D$ is not, in fact, an infinite binary sequence with only finitely many $1$s, do you see why we do not get to conclude that $\{s_n\}$ is incomplete?

Let me reiterate that this answer is not a proof that the set of all infinite binary sequences with finitely many $1$s is countable; in order to prove that it is countable, you need to do some more work (see the other answers, for instance). But hopefully this answer clarifies your intuition about the diagonal argument.
A: Diagonalisation doesn't apply to the set of sequences of $0$s and $1$s with only finitely many $1$s because if you change all the trailing $0$s to $1$s you get a sequence with infinitely many $1$s. A simple way to see why the set is countable is to read a sequence of $0$s and $1$s as a number written in binary notation backwards. For example, read $10110000\ldots$ as $1 + 4 + 8 =  13$. This gives you a one-to-one correspondence between these sequences and the natural numbers.
A: Let $A_n$ be the set of all infinite sequence with $n$ $1$'s and $N_m$ be the set of $n$ $1$'s in first $m$ digits, where $m\geqslant n$. Then $|N_m|$ is finite. Since 
$$
A_n\subset\bigcup_{m=1}^{\infty}N_{m,n}\quad\text{and }\quad T=\bigcup_{n=1}^{\infty}A_n
$$
Clearly $A_n\leqslant \sum_{m\geqslant 1}N_m<\infty$. This means that  $A_n$ is countable.
$T$ is countable for countable union of countable set is countable. 
A: To explicitly list them, note that each such sequence must have a final $1$. For any given $n$, there are finitely many (specifically, $2^{n-1}$) sequences whose final $1$ appears in the $n$th position, so we can list all of them by running through values of $n$ (I will leave off trailing zeroes):

First, the all-zero sequence. Then,
  $n = 1$: $1$
  $n = 2$: $01, 11$
  $n = 3$: $001, 011, 101, 111$
etc.

A: Use the fact that "a countable union of countable sets is countable," several times.


*

*The set of binary strings with finitely many 1s is the union of binary strings with 1 1, 2 1s, 3 1s, 4s, etc.

*The set of binary strings with $n$ 1s is the union of binary strings with $n$ 1s where the greatest 1 is in the $k$ position, the $k+1$ position, the $k+2$ position, etc.

*The set of binary strings with $n$ 1s before the $k$ position and the rest all $0$s is quite clearly less than $2^{k+1}$. Congratulations, you are done.


On further thought you can skip step (1) entirely, but I thought I would include it since it really was the first simplification I thought to make, and you should ruthlessly simplify using the pretty powerful theorem that countable union of countable sets is countable.
A: For $b=(b_n)_{n\in N}\in T,$ if $b_n=0$ for all $n,$  then $f(b)=0.$ 
If $m_0$ is the least $n$ such that $b_n=1,$ and $m_1$ is the largest $n$ such that $b_n=1$  then $$f(b)=(2^{m_0-1})\sum_{j=0}^{m_1-m_0} b_{m_1-j} 2^j.$$ Then $f$ is a bijection from T to the non-negative integers. 
The idea is to interpret $(b_{m_0},...,b_{m_1})$ as the base-two representation of an odd number $g(n)\in N,$ and let $ 2^{m_0-1} g(n)=f(n).$
An attempt to use a diagonal argument to try to prove that $T$ is uncountable will produce a binary sequence with infinitely $1$'s, which is not in $T$.
