Applying the generalized binomial expansion yields $$(1+x)^{\frac{1}{2}}=\sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^n=1+\sum_{n=1}^{\infty}\binom{\frac{1}{2}}{n}x^n$$

Now the coefficient can be transformed as $$\begin{align*} \binom{\frac{1}{2}}{n}=\ &\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots\left(\frac{1}{2}-(n-1)\right)}{n!}\\=\ &\frac{(-1)^{n-1}}{2^n}\cdot\frac{1\cdot 3\cdot 5\cdots (2n-3)}{n!}\\=\ &\frac{(-1)^{n-1}}{2^n\cdot n!}\cdot\frac{(2n-2)!}{2\cdot 4\cdot 6 \cdots  (2n-2)}\\=\ &\frac{(-1)^{n-1}}{2^n\cdot n!}\cdot\frac{(2n-2)!}{(2\cdot 1)(2\cdot 2)(2\cdot 3) \cdots (2\cdot(n-1))}\\=\ &\frac{(-1)^{n-1}}{2^n\cdot n!}\cdot\frac{(2n-2)!}{2^{n-1}\cdot (n-1)!}\\=\ &\frac{(-1)^{n-1}}{2^{2n-1}\cdot n}\cdot\frac{(2n-2)!}{(n-1)!\cdot (n-1)!}\\=\ &\frac{(-1)^{n-1}}{2^{2n-1}\cdot n}\binom{2n-2}{n-1} \end{align*}$$

Setting this result back to the binomial expansion leads to the conclusion immediately.