Yes, your counter-example is allowed. The topology you are describing is the Discrete Topology on $[0,1]$. In this topology, any subset is both open and closed, so in particular the closure of any subset is itself, hence the only dense subset is the entire space. As you note, there are, however, proper subsets of full measure.
For the type of conclusion you want, the measure on your space should have something to do with the topology on your space. To this end, let $X$ be any topological space. The Borel algebra $\mathcal{B}(X)$ is the $\sigma$-algebra on $X$ generated by all the open sets. A measure $\mu$ on $\mathcal{B}(X)$ is called a Borel measure.
Claim: For a topological space $X$ and a Borel measure $\mu$ on $X$, the properties "any Borel set of full measure is dense" and "any non-empty open set has positive measure" are equivalent.
Proof:"$\Rightarrow:$" Assume $U\neq\emptyset$ is an open subset of $X$ and $\mu(U)=0$. Then $\mu(X\setminus(X\setminus U))=\mu(U)=0$, so $X\setminus U$ has full measure. Since $U$ is open, $X\setminus U$ is closed, hence $\overline{X\setminus U}=X\setminus U$, but $X\setminus U\neq X$, since $U\neq\emptyset$, so $X\setminus U$ is not dense in $X$. The claim follows by contrapositive.
"$\Leftarrow:$" Assume $A\subseteq X$ is Borel and not dense in $X$, i.e. $\overline{A}\neq X$. Then $X\setminus\overline{A}\neq\emptyset$. Since $\overline{A}$ is closed, $X\setminus\overline{A}$ is open, so, by hypothesis, $\mu(X\setminus\overline{A})>0$. By definition, $\overline{A}$ does not have full measure. The claim follows by contrapositive.
Note that the second property is also known as strict positivity. In particular, this property holds for every Haar measure, such as the Lebesgue measure on any $\mathbb{R}^n$.
