Implications of compactness in $A\subseteq B\subseteq E$

Question

Let $(E,\tau)$ be a topological space and $A\subseteq B\subseteq E$. Consider the following claims

1. $A$ is compact in $E$
2. $A$ is compact in $B$
3. $B$ is compact in $E$

Since $\left.\left.\tau\right|_B\right|_A=\left.\tau\right|_A$ it should hold (1.) $\Rightarrow$ (2.). On the other hand, if $A$ is closed and $E$ is Hausdorff, it should hold (3.) $\Rightarrow$ (1.). Are the any other implications which are true?

Remark: And just to be sure: The terminology (1.) does mean nothing else than $(A,\left.\tau\right|_A$) is compact and (2.) does mean that $(A,\left.\left.\tau\right|_B\right|_A)$ is compact, right?

Answer 1

$(1)\implies (2)$.

$(2)\not\implies (3)$.
Let $E=\Bbb{R}$, $B=(0,1)$ and take $A$ any finite subset of $B$.

$(3)\not\implies (1)$.
Let $E=\Bbb{R}$, $B=[0,2]$ and $A=(0,1)$.

$(1)\not\implies (3)$.