This may have a simple answer, but I'm not getting anywhere.

If $U$ is an open set in $\mathbb{R}^n$ (usual topology), and $p:[0,1] \to U$ is a continuous path, from $x=p(0)$ to $y=p(1)$, with $x,y \in \mathbb{R}^n$, then can we find an $\epsilon \in (0,1)$ such that $\cup_{t \in [0,1]} B(p(t), \epsilon) \subseteq U$?

Obviously, for any given $t \in [0,1]$, there exists an $\epsilon_t \in (0,1)$ such that $B(p(t), \epsilon_t) \subseteq U$, because $U$ is open. Also we could also take $\epsilon_t = \sup(\{ r \in (0,1) ; B(p(t), r) \subseteq U \})$, so then I'm tempted to take $\epsilon = \inf(\{ \epsilon_t ; t \in [0, 1] \})$. But it's not clear to me how to prove that this last expression can't be zero. This does work if the map $t \mapsto \epsilon_t$ is continuous, but I can't see how to prove that either.

Any help/insight would be appreciated.

Thanks