Can I use Cantor's diagonal argument to show $\{0,1\}^{\mathbb{N}}$ is not countable? So the set can be rewritten as:
$$\{0,1\}^{\mathbb{N}}=\{(a,b,c,d,....) : a,b,c,d,... \in (0,1) \}$$
And so we can write those sequences as:
$$111111111...$$
$$010101010...$$
etc.
And then we pick the $1$st number of the first sequence and change it to $0$, we pick the second of the second etc. etc. and so we will end up with a sequence that is not in any of the sequences in the set, and so $\{0,1\}^{\mathbb{N}}$ is uncountable.
And then a follow up a question. I do not necessarily have to pick the "diagonal" numbers, do I? I can just pick every first number of every sequence and I would still end up with a sequence that is different to every sequence from which it is constructed, correct?
 A: The answer to your first question is yes.
The answer to your second question, however, is no: in order to conclude that some sequence $D$ is not on your list of sequences $S_1 S_2, . . . $, you need to argue:

For each $i$, $D$ disagrees with the $i$th sequence somewhere: that is, for some $n$, $S_i(n)\not=D(n)$.

(This is just because this is what it means for two sequences to be different! Two sequences are different iff they disagree somewhere; so if I want to conclude that $D\not=S_i$ for any $i$, I have to argue that - for each $i$ - there is some "point of disagreement".)
The diagonal argument picks $n=i$: the $i$th bit of $D$ is different from the $i$th bit of $S_i$, for each $i$. We could pick differently: e.g. make the $i+17$th bit of $D$ different from the $i+17$th bit of $S_i$.
By contrast, what you've described - making the $i$th bit of $D$ different from the $1$st bit of $S_i$ - isn't enough! There's no reason to believe that this actually produces a sequence not on the list.
For a concrete example of how this can go wrong, suppose your list looks like:


*

*$S_1=0111111111...$

*$S_2=1000000000...$

*$S_2=1100000000...$

*$S_3=1110000000...$

*And so on.
Then if we do the "change the first bit" trick you describe, the sequence we'll produce is $$10000000000...,$$ which is on the list as $S_2$.
By contrast, the classically-constructed (anti)diagonal sequence is $$11111111111...,$$ which isn't on the list (we know it can't be $S_i$ since its $i$th bit isn't the $i$th bit of $S_i$). Of course, this sequence is in a sense the "limit" of the list, but that's not relevant - it's not on the list.
A: To your first question, yes, you don't have to pick the diagonals, but to your second, no, the new sequence should differ from all those listed in at least one entry.
A: Associate a sequence $s\in\left\{ 0,\,1\right\}^\mathbb{N}$ with $S_s:=\left\{ n\in\mathbb{N}|s_n=1\right\}$. A list $\left( s^j\right) _{j\in\mathbb{N}}$ of such sequences is now associated with $S_{s^j}$ (we use superscripts to number sequences and subscripts to number entries within them). If $\left\{ n\in\mathbb{N}|s^n_n=0\right\}=S_{s^k}$, $s^k_k=0$ iff $k\in S$, i.e. iff $s^k_k=1$.
Depending on how you read this proof by contradiction, you can consider it either the "diagonal argument" on sequences or a special case of the proof of Cantor's theorem (i.e. the result that taking the power set obtains a greater cardinality). Just as one needs to construct a certain set to prove Cantor's theorem, one needs to construct a certain sequence to prove the $s^j$ aren't exhaustive. This relation between subsets and sequences on $\left\{ 0,\,1\right\}$ motivates the description of the proof of Cantor's theorem as a "diagonal argument".
