Give $\Bbb N$ the discrete metric:
$$d(m,n)=\begin{cases}1,&\text{if }m\ne n\\0,&\text{if }m=n\end{cases}$$
Clearly this space is not compact, but any positive $d\le 1$ is a Lebesgue number for every open cover of it.
Added: Having given the matter a bit more thought, I can prove the following theorem. Say that a metric space $\langle X,d\rangle$ is Lebesgue if every open cover of it has a Lebesgue number.
Theorem: Let $\langle X,d\rangle$ be a metric space. If $X$ has a non-convergent Cauchy sequence or an infinite closed discrete set of non-isolated points, then $X$ is not Lebesgue. In particular, every Lebesgue space is complete, and every perfect Lebesgue space is compact.
Proof: Suppose first that $\sigma=\langle x_k:k\in\omega\rangle$ is a non-convergent Cauchy sequence in $X$. Let $\langle X^*,d^*\rangle$ be the usual metric completion of $\langle X,d\rangle$, and let $p\in X^*$ be the limit of $\sigma$ in $X^*$. Let $V_0=X^*\setminus B_{d^*}(p,2^{-1})$, and for $k>0$ let $V_k=B_{d^*}(p,2^{-k+1})\setminus \operatorname{cl}_{X^*}B_{d^*}(p,2^{-k-1})$. For $k\in\omega$ let $W_k=X\cap V_k$. Then $\mathscr{W}=\{W_k:k\in\omega\}$ is an open cover of $X$ with no Lebesgue number.
Now suppose that $\{x_k:k\in\omega\}$ is a closed discrete set of non-isolated points in $X$. There is a pairwise disjoint, closure-preserving collection $\{V_k:k\in\omega\}$ such that $x_k\in V_k$ for each $k\in\omega$, so there is a sequence $\langle r_k:k\in\omega\rangle$ of positive real numbers such that $B_d(x_k,r_k)\subseteq V_k$ for each $k\in\omega$, and $\langle r_k:k\in\omega\rangle\to 0$. Let $$W=X\setminus\bigcup_{k\in\omega}\operatorname{cl}_X B_d\left(x_k,\frac{r_k}2\right)\;,$$ and let $\mathscr{W}=\{W\}\cup\{B_d(x_k,r_k):k\in\omega\}$; then $\mathscr{W}$ is an open cover of $X$ with no Lebesgue number. $\dashv$
That set is not compact. Consider the open sets$$\left\{(x,y)\in\mathbb{R}^2\,\middle|\,\bigl\lVert(x,y)-(1,0)\bigr\rVert>\frac1n\right\},$$with $n\in\mathbb N$. These sets form an open cover of your set without a finite subcover.
Best Answer
The answer to your question is yes. In a metric space $X$, $X$ is open. Since (very reduntantly) every subset of $X$ is a subset of $X$, then $X$ functions as an open cover for each of its subsets. This observation is not useful though, and has nothing to do with compactness. A set $K\subset X$ of a metric space $X$ is compact iff every open cover of $K$ has a finite subcover. An open cover of $K$ could contain finitely many open sets, or it could contain infinitely many open sets. In the latter case, you have to know that you can choose finitely many open sets out of the infinite collection and still be able to cover $K$. If you cannot, then $K$ isn't compact.
It might be helpful to work through an example. I'm sure it's in your textbook, but can you find an open cover of $(0,1)$ in the metric space $(\Bbb{R},d)$ [where $d$ is the Euclidean distance metric] that does not have a finite subcover?