Trying to understand the proof of Lemma 1.15 in Hatcher’s Algebraic Topology Ok so I’m currently going through Hatcher’s Algebraic Topology, and in the first chapter he proves what turns out to be a very important Lemma (see below) and the proof is actually fairly straightforward but there are a few points that I’m not entirely sure I understood. The lemma is the following:

Lemma 1.15 If a space $X$ is the union of a collection of path-connected open sets $A_\alpha$ each containing the basepoint $x_0\in X$ and if each intersection $A_\alpha \cap A_\beta$ is path-connected, then every loop in $X$ at $x_0$ is homotopic to a product of loops each of which is contained in a single $A_\alpha$.

As I said, I understand the proof, there’s just one detail that I’d like to clarify. Namely the very first step of the proof is to exhibit a certain partition of $I$ (unit interval) such that the loop $f$ maps each sub interval of the partition to a single $A_\alpha$. Hatcher writes

Since $f$ is continuous, each $s\in I$ has an open neighbourhood $V_s$ in $I$ mapped by $f$ to some $A_\alpha$.

This seems obvious but actually upon thinking about it I wasn’t sure anymore. My guess at showing that is maybe something like:
Take $s\in I$ then by definition $f(s)$ lies in some $A_\alpha$. Let $N$ be an open neighbourhood of $f(s)$ in $A_\alpha$, by continuity $f^{-1}(N)$ is open in $I$ so take $V_s = f^{-1}(N)$ and that’s it. Is that correct ?
Furthermore, the next sentence in the proof is

We may in fact take $V_s$ to be an interval whose closure is mapped to a single $A_\alpha$.

This is a little less obvious but intuitively I just thought that if you take a neighbourhood of $f(s)$ small enough at some point $V_s$ ends up being a single open interval and not a union of open intervals but this isn’t an argument, rather a gut-feeling and I’ve been thinking about it and unsure how to really convince myself that this is true.
So if anyone is willing to clarify those points for me, I’d be grateful, thank you.
 A: The set $A_\alpha$ is an open subset of $X$, and the function $f : I \to X$ is continuous. By applying the definition of continuity, it follows that the set $f^{-1}(A_\alpha)$ is an open subset of $I = [0,1]$.
Since $s \in f^{-1}(A_\alpha)$, by applying the definition of the topology on $[0,1]$ there exists an open neighborhood $W_s$ of $s$ that is entirely contained in $f^{-1}(A_\alpha)$, and that has one of the following forms (these are neighborhood basis elements around $s$ in the topology on $I$):

*

*if $s$ is in the interior of $I = [0,1]$, namely if $s \in (0,1)$, then $W_s = (s-r,s+r)$;

*if $s = 0$ then $W_s = [0,r)$

*if $s=1$ then $W_s = (1-r,1]$.

You can then define $V_s$ to be the corresponding open neighborhood of half radius $r/2$:  $(s-r/2,s+r/2)$  or $[0,r/2)$ or $(1-r/2,1]$.
The closure of $V_s$ is therefore either $[s-r/2,s+r/2]$ or $[0,r/2]$ or $[1-r/2,1]$, which is a subset of $W_s$, which is a subset of $f^{-1}(A_\alpha)$, which is mapped to $A_\alpha$ by the function $f$.
