Let $\mathcal{B}\subset\mathcal{A}$ be an inclusion of unital $C^*$-algebras. Let $b\in \mathcal{B}$ be a non-negative non-invertible element.
Question: Is it possible to find an element $a\in\mathcal{A}\setminus\mathcal{B}$ such that $0<a<b$?
Of course, we can always find an element $a'\in\mathcal{B}$ by just taking $a'=\epsilon b$ for $\epsilon<1$.
Edit: Can we find $b\in\mathcal{B}$ and $a\in\mathcal{A}\setminus\mathcal{B}$ (depending on $b$) such that $0<a<b$?
No, take $A$ to be any unital algebra and $B$ to be hereditary in $A$: by definition, this means that the positive cone of $B$ is hereditary in the positive cone of $A$, that is, if $0\le a\le b$ and $b\in B$, then $a\in B$. Obviously, if $B$ is hereditary in $A$ your question gets a negative answer. Now examples of hereditary inclusions are in abundance; for example, if $A$ is any $C^*$-algebra and $p\in A$ is a projection, then $pAp$ is a hereditary $C^*$-subalgebra of $B$. In particular, if $X=X_1\sqcup X_2$ is a compact Hausdorff space with two connected components $X_1,X_2$, then $C(X_i)\subset C(X)$ is a hereditary inclusion (apply the above with $A=C(X)$ and $p$ the indicator function of $X_i$).