Lemma. If $A$ and $B$ are nonempty sets, and there is a one-to-one function $f\colon A\to B$, then there is a surjective function $g\colon B\to A$.
Proof. Let $a_0\in A$ (possible, since $A$ is nonempty). Define $g\colon B\to A$ as follows:
$$g(b) = \left\{\begin{array}{ll}
a_0 &\text{if }b\notin f(A);\\
a &\text{if }b\in f(A)\text{ and }f(a)=b.
\end{array}\right.$$
Note that $g$ is well defined, since $f$ is one-to-one, so there is at most one $a\in A$ with $f(a)=b$. And $g$ is onto, because give $a\in A$, $g(f(a))=a$. $\Box$
Now, we are assuming that there exists an embedding $(0,1)\hookrightarrow A$. Therefore, there is a surjection $g\colon A\to (0,1)$. If $f\colon\mathbb{N}\to A$, then $g\circ f\colon \mathbb{N}\to (0,1)$ is a function that is not onto. Conclude that $f$ is not onto. Conclude that $A$ is not countable.
Alternatively, by contradiction: suppose $f\colon\mathbb{N}\to A$ is onto. Let $S\subseteq \mathbb{N}$ be the set of all natural numbers such that $f(n)\in (0,1)$ (we are vieweing $(0,1)$ as a subset of $A$ via the inclusion). Then $f$ would be an onto function from a countable set (every subset of $\mathbb{N}$ is countable) onto $(0,1)$, which contradicts the fact that $(0,1)$ is not countable.