I'd like to prove the following result, perhaps with additional assumption if needed -- I don't know whether the claim holds. Let $M$ be a compact connected manifold with boundary $\partial M$. Let $U' \subset \partial M$ be open in $\partial M$. Then there exists $U \subset M$ open in $M$ such that $U \cap \partial M = U'$, and $U'$ is a strong deformation retract of $U$. Any ideas welcome.
Background
This result is true in the following specific case which I'd like to generalize. Let $M = \overline{B_n}$ be the closed unit n-ball at origin in $\mathbb{R}^n$. Then $\partial M = S^{n - 1}$, the unit $(n-1)$-sphere at origin. Let $U' \subset S^{n - 1}$ be open in $S^{n - 1}$. Let $U = \{(1 - t)u' : t \in [0, 1/2) \text{ and } u' \in U'\}$. Then $U$ is open in $M$ and $U \cap \partial M = U'$. Let $f : U \times [0, 1] \to U$ be such that $f(x, t) = (1 - t)x + tx/|x|$. Then $f$ is a strong deformation retraction of $U$ onto $U'$.
There is this thing called collar neighbourhood of $\partial M$. It's an open neighbourhood of $\partial M$ in $M$ which is homeomorphic to $\partial M\times [0,1)$ and the homeomorphism maps $x\mapsto (x,0)$ for $x\in\partial M$. In fact you did use a collar neighbourhood in the specific case of $n$-ball.
So if $U'\subseteq\partial M$ is open and $\partial M$ has a collar neighbourhood then we can restrict the collar neighbourhood to $U'\times[0,1)$ which is again open in $M$ because it is open in $\partial M\times [0,1)$. So we obtain an open subset of $M$ homeomorphic to $U'\times[0,1)$. And such neighbourhood easily (strongly) deformation retracts onto $U'$.
Compact manifolds are guaranteed to have collar neighbourhoods. You can find the proof in Allen Hatcher's "Algebraic Topology", Proposition 3.42.
Edit: According to Morton Brown, "Locally flat imbeddings of topological manifolds" every manifold with boundary has a collar neighbourhood. (thanks @MoisheKohan for noticing)