How do I evaluate $\prod_{r=1}^{\infty }\left (1-\frac{1}{\sqrt {r+1}}\right)$?

The product is $0$ (therefore by the definition of a converging infinite product your product does not really converge). For every integer $n\geq 1$ we have \begin{equation*} 0 \,<\, \prod_{r=1}^n\left(1-\frac{1}{\sqrt{r+1}}\right) \,\leq\, \exp\left(-\sum_{r=1}^n\frac{1}{\sqrt{r+1}}\right)~. \end{equation*} The series $\sum_{r=1}^\infty 1/\sqrt{r+1}$ diverges, so the partial products of your infinite product tend to $0$.

If you know that, for $0<a_n<1$, we have $$\prod_{i=1}^{\infty} (1-a_n) = 0 \text{ if and only if } \sum_{i=1}^{\infty} a_n = \infty$$ then it suffices if you show that $$\sum_{r=1}^\infty \frac1{\sqrt{r+1}} = +\infty.$$

For the proof of the above general fact, see How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$?

For the series which you get see, for example, Is $\sum\frac{1}{\sqrt{n+1}}$ convergent or divergent?