By the generalized binomial theorem,
$$
(1-z)^{-1/2}=
\sum_{k=0}^\infty\binom{-1/2}{k}(-z)^k
$$
so your task is to prove that
$$
(-1)^k\binom{-1/2}{k}=\frac{1}{2^{2k}}\binom{2k}{k}
$$
This is surely true for $k=0$. Next, by induction,
$$
(-1)^{k+1}\binom{-1/2}{k+1}=
-(-1)^k\binom{-1/2}{k}\frac{-1/2-k}{k+1}=
\frac{1}{2^{2k}}\binom{2k}{k}\frac{1}{2}\frac{2k+1}{k+1}
$$
Personally, I wouldn't have done it that way. So here is how I would've done it:
Method 1:
$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$
which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.
$$1.375^2=1.890625$$
Obviously it approaches the correct value as you take more terms.
Method 2:
This is called fixed-point iteration/Newton's method, and it basically goes like this:
$$x=\sqrt2\implies x^2=2$$
$$2x^2=2+x^2$$
Divide both sides by $2x$ and we get
$$x=\frac{2+x^2}{2x}$$
Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so
$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$
and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$. For example, with $x_0=1$,
$x_0=1$
$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$
$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$
$x_3=\dots=\frac{577}{408}=1.414215686$
And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.
Best Answer
Here we look at some arguments why taking out a factor $2.5$ is a convenient choice.
In order to apply (1) we are looking for a number $y$ with \begin{align*} \sqrt{1-4x}&=\sqrt{6y^2}=y\sqrt{6}\tag{2}\\ \color{blue}{\sqrt{6}}&\color{blue}{=\frac{1}{y}\sqrt{1-4x}} \end{align*}
We see it is convenient to choose $y$ to be a square number which can be easily factored out from the root. We obtain from (2) \begin{align*} 1-4x&=6y^2\\ 4x&=1-6y^2\\ \color{blue}{x}&\color{blue}{=\frac{1}{4}-\frac{3}{2}y^2}\tag{3} \end{align*}
We see in (1) a good approximation is given if $x$ is close to $0$. If $x$ is close to zero we will also fulfill the convergence condition. $x$ close to zero means that in (3) we have to choose $y$ so that \begin{align*} \frac{3}{2}y^2 \end{align*} is close to $\frac{1}{4}$. So, $y^2$ is close to $\frac{1}{4}\cdot\frac{2}{3}=\frac{1}{6}$. We have already (1) and (3) appropriately considered. Now we want to find small natural numbers $a,b$ so that
\begin{align*} y^2=\frac{a^2}{b^2}\approx \frac{1}{6} \end{align*} This can be done easily. When going through small numbers of $a$ and $b$ whose squares are apart by a factor $6$ we might quickly come to $100$ and $16$. These are two small squares and we have $6\cdot 16=96$ close to $100$. That's all.
At this point it might be clear that factoring out $2.5=\frac{5}{2}=\frac{1}{y}$ is a good choice.
Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.