In $\mathbb{R}^n$ with the usual metric, "compact" is equivalent to "bounded and closed" (Heine-Borel-Lebesgue). However, the equivalence does not hold in arbitrary metric spaces.
The correct statement for arbitrary metric spaces is:
Theorem :
Let $(X,d)$ be a metric space. A subset $S$ of $X$ is compact if and only if it is complete and totally bounded.
Recall that a set $S$ is "complete" if every Cauchy sequence in $S$ converges; and is "totally bounded" if for every $\epsilon\gt 0$ there exists a finite number of open balls of radius $\epsilon$ that cover $S$.
In $\mathbb{R}^n$ with the usual topology, totally bounded and bounded are equivalent, as are "complete" and "closed".
However, for example, if you change the metric to $d(x,y)=\arctan|x-y|$, then the space is bounded but not totally bounded. So,
EDIT
To see that $\mathbb{R}$ is bounded but not totally bounded under $d$, note that $0\leq\arctan|x-y|\lt\frac{\pi}{2}$, so the space is bounded, contained in the ball of radius $\frac{\pi}{2}$ centered at $0$. To see that $\mathbb{R}$ is not totally bounded under $d$, pick $\epsilon\gt 0$ that is very small. Then $y\in B_d(x,\epsilon)$ if and only if $\arctan|x-y|\lt\epsilon$, if and only if $|x-y|\lt\tan(\epsilon)$. For $\epsilon$ very close to $0$, we have $\tan(\epsilon)\approx\epsilon$, so these balls are "essentially" the same size they would be in the Euclidean metric; for very small $\epsilon\gt 0$, then, you cannot cover $\mathbb{R}$ with finitely many $\epsilon$-$d$-open balls, so $\mathbb{R}$ is not totally bounded under $d$.
So you cannot make the closed set $A_1$ compact merely by replacing the metric with an equivalent-but-bounded metric, because this exchange does not respect the property of being totally bounded.
For example in the usual metric where $A_k$ is and nonempty and closed for all $k$, but the intersection is empty, just take $A_n=[n,\infty)$ in $\mathbb{R}$. Easily generalized to $\mathbb{R}^n$ with the Euclidean metric.