I want to show that
for all nonemptyset $A\subset \mathbb R$, there exists a sequence $\{a_n\}_{n=1}^\infty\subset A$ s.t. $\lim a_n=\inf A\ \ $ $\cdots(\ast)$
I know the fact that if $\{b_n\}_{n=1}^\infty\subset \mathbb R$ is decreasing, then $\lim b_n=\inf \{b_n\}_{n}$.
I think this is similar to $(\ast)$ so I can perhaps use this fact to prove $(\ast)$.
But to do so, I have to pick a decreasing sequence $\{a_n\}$ in $A$, and I have diffuculty in finding such $\{a_n\}$.
Thanks for any help. Another way to prove is also welcomed.
Your claim is wrong as stated because not every nonempty set in $\mathbb{R}$ has an infimum. However, every nonempty set which is bounded below does have an infimum and the claim is true in that case. To see that: let's call $x = \inf A$. By definition of infimum, there exists $a_1 \in A$ such that $a_1 < x + \frac{1}{2}$. Similarly, for every $i \in \mathbb{N}$ there exists an $a_i \in A$ (not necessarily distinct from one another, they could all be the same element of $A$, or the sequence could be eventually constant) such that $a_i < x + 2^{-i}$. It is clear then that $\displaystyle{\lim_{n \in \mathbb{N}} a_n \leq x} $. Again by the definition of infimum it follows that $\displaystyle{\lim_{n \in \mathbb{N}} a_n} = x$, so you have the desired sequence. Exercise: adapt this argument to show a similar statement for the supremum of sets bounded above.
Some more requested details: Since $x$ is the infimum of $A$, it follows that $a_i -x \geq 0$ for all $i \in \mathbb{N}$. By construction we also have $a_i - x < 2^{-i}$ for al $i \in \mathbb{N}$. Therefore $|a_i - x| < 2^{-i}$ for all $i \in \mathbb{N}$. It follows from the archimedean property of $\mathbb{R}$ that for any $\varepsilon > 0$, there exists $n_0 \in \mathbb{N}$ such that $|a_k - x| < \varepsilon$ for all $k \geq n_0$. By definition, it is then clear that $\displaystyle{\lim_{n \in \mathbb{N}} a_n}$ exists and is equal to $x$.