Applying the generalized binomial theorem gives $$\frac{1}{\sqrt{1-4x}}=(1-4x)^{-\frac{1}{2}}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}(-1)^n2^{2n}x^n$$

The coefficient can be simplified as $$\begin{align*}\binom{-\frac{1}{2}}{n}(-1)^n2^{2n}=\ &\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!} (-1)^n2^{2n}\\=\ &\frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!}\cdot 2^n\\=\ &\frac{(2n)!}{n!(2\cdot 4\cdots (2n))}\cdot 2^n\\=\ &\frac{2n!}{n!}\frac{1}{(2\cdot 1)(2\cdot 2)\cdots (2\cdot n)}\cdot 2^n\\=\ &\frac{(2n)!}{n!n!}\\=\ &\binom{2n}{n}\end{align*}$$

Setting this back to the first equation leads to the claim immediately.