Let
$$
f(x) = \frac{x}{x^2-1}.
$$
This is a homeomorphism from $(-1,1)$ to $\mathbb R$.
Let
$$
g(x) = \frac{1}{1+2^{-x}}.
$$
That is a homeopmorphism from $\mathbb R$ to $(0,1)$.
For any other bounded interval $(a,b)$, just rescale and relocate.
You get the uniqueness result if the space is Hausdorff.
Let $\langle X,\tau\rangle$ be a compact space. Suppose that $p\in X$ is in the closure of $Y=X\setminus\{p\}$, and let $\tau_Y$ be the associated subspace topology on $Y$; $\langle X,\tau\rangle$ is then a compactification of $\langle Y,\tau_Y\rangle$.
Suppose that $p\in U\in\tau$, and let $V=U\cap Y$. Then $\varnothing\ne V\in\tau_Y$, so $Y\setminus V$ is closed in $Y$. Moreover, $Y\setminus V=X\setminus U$ is also closed in $X$, which is compact, so $Y\setminus V$ is compact. That is, every open nbhd of $p$ in $X$ is the complement of a compact, closed subset of $Y$. Thus, if $\tau'$ is the topology on $X$ that makes it a copy of the Alexandroff compactification of $Y$, then $\tau\subseteq\tau'$.
Now let $K\subseteq Y$ be compact and closed in $Y$, and let $V=Y\setminus K\in\tau_Y$. If $X\setminus K=V\cup\{p\}\notin\tau$, then $p\in\operatorname{cl}_XK$. If $X$ is Hausdorff, this is impossible: in that case $K$ is a compact subset of the Hausdorff space $X$ and is therefore closed in $X$. Thus, if $X$ is Hausdorff we must have $\tau=\tau'$, and $X$ is (homeomorphic to) the Alexandroff compactification of $Y$.
If $X$ is not Hausdorff, however, we can have $\tau\subsetneqq\tau'$. A simple example is the sequence with two limits. Let $D$ be a countably infinite set, let $p$ and $q$ be distinct points not in $D$, and let $X=D\cup\{p,q\}$. Points of $D$ are isolated. Basic open nbhds of $p$ are the sets of the form $\{p\}\cup(D\setminus F)$ for finite $F\subseteq D$, and basic open nbhds of $q$ are the sets of the form $\{q\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $Y=D\cup\{q\}$. Then $Y$ is dense in $X$, and $X$ is compact, and $Y$ itself is a closed, compact subset of $Y$ whose complement is not open in $X$.
Improved example (1 June 2015): Let $D$ and $E$ be disjoint countably infinite sets, let $p$ and $q$ be distinct points not in $D\cup E$, let $X=D\cup E\cup\{p,q\}$, and let $Y=D\cup E\cup\{q\}$. Points of $D\cup E$ are isolated. Basic open nbhds of $q$ are the sets of the form $\{q\}\cup (E\setminus F)$ for finite $F\subseteq E$, and basic open nbhds of $p$ are the sets of the form $\{p\}\cup\big((D\cup E)\setminus F\big)$ for finite $F\subseteq D\cup E$. Then $Y$ is a non-compact dense subspace of the compact space $X$, so $X$ is a (non-Hausdorff) compactification of $Y$. Let $K=\{q\}\cup E$. Then $K$ is a compact closed subset of $Y$, but $X\setminus K=\{p\}\cup D$ is not open in $X$.
(This avoids the question of whether it’s legitimate to look at the Alexandrov compactification of a compact space.)
Best Answer
Every open interval $(a,b)\subseteq \mathbb R$ is indeed homeomorphic to $\mathbb R$. The one-point compactification of $\mathbb R$ is homeomorphic to the unit circle $\mathcal S^1$.
Now, $\mathbb R$ and $\mathcal S^1$ are not homeomorphic because the latter is compact (and the real line isn't). Is that what you mean by "compact circumference"?
The two-point compactification of $\mathbb R$ is homeomorphic to the closed interval $[0,1]$ (or any other closed real interval).
So it looks like you got it right.
[Also, here's a post where you can find a definition of the two-point compactification and also an interesting discussion: Question on compactification ]