Why is the collection of all infinite sequence coin tosses uncountable? NOTE: The question may appear unclear due to my very limited knowledge about the subject.
So I my instructor something along the lines that if we have collection of infinite sequence of coin tosses and assign a binary ${0.1}$ to Heads and Tails we can create a one to one correspondence between the collection of sequences and $[0,1]$ (which is uncountably infinite).
How is this possible? Since real line contains rational numbers, while here we are summing up rational numbers, how can we come up with one to one correspondence with irrational numbers? This is considering that rational numbers does not have the least upper bound property i.e. roughly there are certain numbers it can never reach, or there are gaps in rational numbers. How to prove that in this case no such gap exists in such infinite collections of binary strings?
 A: There is a $1-1$ correspondence between the real numbers in the range $[0,1]$ and the infinite binary strings.
We just have to add "$0.$" before the infinite bit string to get a real number in this range in binary expansion and conversely , "$0.$" returns the corresponding string.
Since the set of real numbers in the range $[0,1]$ is uncountable, we are done.
A: Let $S=\big\{(a_n)_{n=1}^\infty:a_n\in\{0,1\}\big\}$ be the set of infinite sequences of coin tosses (so for example assign $0$ and $1$ to "heads" and "tails")
Claim 1: If $X$ is a countable subset of $S$ then $S\backslash X$ is nonempty.

Proof: Suppose $X\subseteq S$ is countable. We may clearly assume $X$ is infinite. So enumerate $X=\{s_1,s_2,s_3,\ldots\}$. Say $s_i=(a_{i,n})_{n=1}^\infty$ where $a_{i,n}\in\{0,1\}$. Define a sequence $s=(a_n)_{n=1}^\infty$ so that $a_n=0$ if and only if $a_{n,n}=1$. Then $s\neq s_i$ for any $i$, since $s$ and $s_i$ disagree at coordinate $i$. So $s\in S\backslash X$.

It follows immediately that $S$ is uncountable. As your instructor says, $S$ has the same cardinality as $[0,1]$. As you've already seen from the extensive discussion, this is usually done explicitly using a bijection between sequences in $S$ and binary representations of real numbers. But, in addition to your qualms about sums of rationals, there are some annoying technicalities having to do with uniqueness of binary representations. So let me first warmup with a different identification of the exact cardinality of $S$.
Let $P(\mathbb{N})$ denote the set of all subsets of $\mathbb{N}=\{1,2,3,\ldots\}$.
Claim 2: $S$ has the same cardinality as $P(\mathbb{N})$.

Proof: Define a function $f:S\to P(\mathbb{N})$ such that, if $s=(a_n)_{n=1}^\infty$ is in $S$, then $f(s)$ is the set $\{n\in\mathbb{N}:a_n=1\}$. It is an elementary exercise that $f$ is one-to-one and onto.

Now it is a well-known fact that $P(\mathbb{N})$ has the same cardinality as $\mathbb{R}$, which has the same cardinality as the interval $[0,1]$ (or $(0,1)$ or $[0,1)$ or any nontrivial interval of real numbers). A very detailed account can be found here.  There are many such sources online.

Remark. We can easily define a 1-1 function from $\mathbb{R}$ to $S$, which gives us another way to see that $S$ is uncountable. Enumerate the rationals $\mathbb{Q}=\{q_1,q_2,q_3,\ldots\}$. Given a real number $r$, define a sequence $s(r):=(a_n)_{n=1}^\infty$ such that $a_n=1$ if and only if $r\leq q_n$. Then $r\mapsto s(r)$ is 1-1 since the rationals are dense.

In order to make an explicit 1-1 correspondence between $S$ and $[0,1]$, the most natural way is to use binary representations.
So let's do it. Given a sequence $s=(a_n)_{n=1}^\infty$, define the real number
$$
r_s=\sum_{n=1}^\infty a_n2^{-n}.
$$
The sum on the right hand side converges to a well-defined number in $[0,1]$ (calculus exercise).
We now have a map from $S$ to $[0,1]$ sending $s$ to $r_s$. Unfortunately this map is not injective. For example $(0,1,1,1,1,\ldots)$ and $(1,0,0,0,0,\ldots)$ are both sent to $1/2$. So to fix this issue, we first focus on sequences that are not eventually constantly $1$.
Define $T$ to be the set of $(a_n)_{n=1}^\infty$ in $S$ such that $a_n=0$ for arbitrarily large $n$.
Lemma 3. $s\mapsto r_s$ is a 1-1 correspondence between $T$ and $[0,1)$.

Proof. Note that the only sequence sent to $1$ is $(1,1,1,1,\ldots)$ which is not in $T$. So $r_s\in [0,1)$ for all $s\in T$.
Now suppose we have distinct sequences $s=(a_n)_{n=1}^\infty$ and $t=(b_n)_{n=1}^\infty$ in $T$. Let $N$ be minimal such that $a_N\neq b_N$. Without loss of generality, $a_N=0$ and $b_N=1$. Then we can write $r_s=u+\sum_{n=N+1}^\infty a_n2^{-n}$ and $r_t=u+2^{-N}+\sum_{n=N+1}^\infty b_n2^{-n}$ for some real number $u$. In particular $r_t\geq u+2^{-N}$. On the other hand $r_s<u+2^{-N}$ since $\sum_{n=N+1}^\infty a_n2^{-(n+1)}<2^{-N}$ (here we use the assumption that the coefficients $a_n$ in the last sum are not all $1$). So $r_s\neq r_t$, and we have shown that our map is 1-1.
Finally we need to show that any real number in $[0,1)$ is of the form $r_s$ for some $s\in T$. So fix $r\in [0,1)$. Pick $a_1\in \{0,1\}$ such that $a_1=0$ if and only if $r<1/2$. Since $r<1$, we get $a_12^{-1}\leq r< 2^{-1}+a_12^{-1}$. Now suppose we have $a_1,\ldots,a_n$ such that
$$
\sum_{i=1}^n a_i2^{-i}\leq r< 2^{-n}+\sum_{i=1}^n a_i2^{-i}.
$$
Define $a_{n+1}\in\{0,1\}$ such that $a_{n+1}=0$ if and only if $r-\sum_{i=1}^n a_i2^{-i}<2^{-(n+1)}$. Then
$$
\sum_{i=1}^{n+1}a_i2^{-i}\leq r<2^{-(n+1)}+\sum_{i=1}^{n+1}a_i2^{-i}.
$$
This inductively constructs a sequence $(a_n)_{n=1}^\infty$ such that, for all $n\geq 1$, $\left|r-\sum_{i=1}^n a_i2^{-i}\right|<2^{-n}$. It follows that $r=\sum_{n=1}^\infty a_n 2^{-n}$.
Now we only need to show that $(a_n)_{n=1}^\infty$ is in $T$. Suppose not. Then there is some $n\geq 1$ such that $a_{n}=0$ and $a_i=1$ for all $i> n$.  So
$$
r=\sum_{i=1}^{n-1}a_i2^{-i}+\sum_{i=n+1}^\infty 2^{-i}=\sum_{i=1}^{n-1}a_i2^{-i}+2^{n}
$$
Then $r-\sum_{i=1}^{n-1}a_i2^{-i}=2^{-n}$, which means we set $a_n=1$ in the construction above. This is a contradiction.

The previous lemma gets us well on our way to building the desired 1-1 correspondence between $S$ and $[0,1]$. However, we have a problem, which is that there are many sequences in $S$ that are not in $T$, but the map $s\mapsto r_s$ for $s\in T$ already uses up everything in $[0,1]$ except $1$. So we have to fiddle with the map a little bit. The main observation is:
Claim 4. $S\setminus T$ is countable.

Proof. This is basically obvious. Any $s\in S\setminus T$ is completely determined by some finite initial segment, and so $S\setminus T$ is in 1-1 correspondence with the set of finite sequences of $0$’s and $1$’s.

So now lets start fiddling with the map. Let $Q=S\setminus T$, which is countable by Claim 4. In order to adjust $s\mapsto r_s$ to account for $Q$, we will just choose a countable subset of $[0,1]$ (namely, $\{\frac{1}{n}:n\geq 1\}$) and do a “Hilbert hotel” shift to make room for $Q$.
In particular, enumerate $Q=\{q_1,q_2,q_3,\ldots\}$. Define $f:S\to [0,1]$ such that
$$
f(s)=\begin{cases}
r_s & \text{if $r_s\not\in Q$ and $r_s\neq \frac{1}{n}$ for all $n\geq 1$}\\
\frac{1}{2n} & \text{if $s\not\in Q$ and $r_s=\frac{1}{n}$}\\
\frac{1}{2n-1} & \text{if $s=q_n$}
\end{cases}
$$
Then $f$ is a 1-1 correspondence by construction.
