It is trivial to obtain, from squaring the binomial expansion of $(1+t)^{1/2}$ and comparing coefficients, that $$\tag1 \sum_{k=0}^m\binom{1/2}k\binom{1/2}{m-k}=0,\qquad m\geq2. $$ This can be rewritten as $$\tag2 \sum_{k=1}^{m-1}\frac{(2k-2)!(2(m-k)-2)!}{k!(k-1)!(m-k)!(m-k-1)!}=\frac{(2m-2)!}{m!(m-1)!},\qquad m\geq2. $$
I wonder if there is a direct proof of these equalities, in either version.
Your equation (2) can be rewritten as $$\sum_{k=1}^{m-1} C_{k-1}C_{m-k-1} = C_{m-1},$$ where $C_n$ is the $n^{\text{th}}$ Catalan number. Any introductory article on Catalan numbers should contain one or more counting proofs of that identity.