Is $(0,1)$ a compact metric space? It is well known that the open interval $(0,1)$ is not compact on the real line. But I somehow remembered that if we take the open interval $(0,1)$ itself as a metric space, then it is a compact metric space. Is this true? Sorry if this question is naive...
 A: Adding a small answer : A set is compact $\iff $ it is complete and totally bounded.
But $(0,1)$ is not complete ;consider the sequence $\{\frac{1}{n}:n\in \mathbb N\}$
A: (I'm going to put this answer in terms of topological spaces.  If you're not comfortable enough with general topological spaces, just replace "topological" with "metric" throughout.  In this case, subspace topology is the same as talking about restricting the metric on $X$ to only $Y$)  
No, given a topological space $X$ and a subspace $Y$, then $Y$ is compact in the subspace topology if and only if it is compact as a subset of $X$, i.e. if for every open cover of $Y$ by open sets in $X$ there is a finite subcover.
Suppose that $Y$ is compact and $\mathcal{A}=\{A_\alpha\}_{\alpha\in J}$ is a covering of $Y$ by sets open in $X$.  Then the collection $\{A_\alpha\cap Y\mid \alpha\in J\}$ is a covering of $Y$ by sets open in $Y$, so that there is a finite subcover $\{A_{\alpha_1}\cap Y,\ldots, A_{\alpha_n}\cap Y\}$ that corresponds to an open cover $\{A_{\alpha_1},\ldots, A_{\alpha_n}\}$ of $\mathcal{A}$.
Conversely, suppose that every open cover of $Y$ of sets open in $X$ admits a finite subcover of $Y$ of sets open in $X$.  Let $\mathcal{A}'=\{A_\alpha\}_{\alpha\in J}$ be an open cover of $Y$ by sets open in $Y$.  Then for each $\alpha$, choose $A_\alpha$ open in $X$ such that $A'_\alpha=A_\alpha\cap Y$.  Then $\mathcal{A}=\{A_\alpha\}$ is a cover of $Y$ by sets open in $X$, and by assumption, there is a finite subcover $\{A_{\alpha_1},\ldots, A_{\alpha_n}\}$ of open sets in $X$, so that $\{A_{\alpha_1}',\ldots, A_{\alpha_n}'\}$ is a finite subcover of $\mathcal{A'}$ of sets open in $Y$.
Applying the above to your problem, we see that because $(0,1)$ isn't compact as a subset of $\mathbb{R}$, it isn't compact when viewed as a topological space in its own right (with the subspace topology).
A: It is not true. Compactness is an topological property which does not depends on the embedding $(0,1)\to \mathbb R$. The inclusion $(0,1)\to \mathbb R$ is only used to give $(0,1)$ a topology (the subspace topology) so that we can talk about compactness. 
That $(0,1)$ under the subspace topology is noncompact is from Heine-Borel Theorem, which state that a set $K$ in $\mathbb R$ is comapct if and only if it is closed and bounded. 
As a set, one can give $(0,1)$ a metric $d$ so that $((0,1),d)$ becomes a compact metric space. For example, let $f:(0,1) \to [0,1]$ be a bijection and define $d(x, y) = |f(x) - f(y)|$. However this metric space has nothing to do with the subspace topology. Indeed there isn't a continuous surjective mapping 
$$((0,1), d) \to ((0,1) , \mathscr T_e)$$
as $((0,1), d)$ is compact and continuous image of compact set is compact ($\mathscr T_e$ is the usual subspace topology of $(0,1)$, treated as a subset of $\mathbb R$). 
A: Don't take this answer too seriously but: 
You know $M=(0,1)$ and $I=[0,1]$ have the same cardinality.
Let $f:M\to I$ be a bijection. Define a metric $\rho$ on $M$ by 
$\rho(p,q)=|f(p)-f(q)|$. Introduce a topology on $M$ using the metric $\rho$ just defined. Then $f$ becomes an isometry as well as a homeomorphism between $M$ and $I$. Since $I=[0,1]$ (with the usual metric $d(x,y)=|x-y|$) is compact, so is $M=(0,1)$ with the metric $\rho$. 
