Simply connected subset of $\mathbb{S}^1$

Question

I was trying. to think of a generalization for theorem stating that if $X=U\cup V$ where $U,V$ are simply connected with path-connected intersection, then $X$ is simply connected.

But I was given the counter-example of covering $\mathbb{S}^1$ by 3 arcs each of length $=\left(\frac{1}{3}+\epsilon\right)*$(circumference of the circle), and I cannot understand how this counter example demonstrates simply connected subsets, or how the same argument couldn't be used for the case of the initial theorem? i.e. Having two arcs of length$=\left(\frac{1}{2}+\epsilon\right)*$(circumference of the circle)?

Edit:

** I am going to include my attempt to prove the generalization I was going for. My limited experience with math has taught me that finding my errors is just as insightful, if not more so, than having the correct solution. So, if someone could point out my incorrect assumption or conclusion at any step, I would be grateful. (Sorry for the length, I try to include all steps, but this should make it very easy to isolate where it went bad)

Statement: Suppose $X$ is a topological space and $\mathcal{U}$ is an open cover of $X$ such that each $U\in \mathcal{U}$ is simply connected, and that the intersection of any 2 sets in $\mathcal{U}$ is path-connected; i.e. for any $U_{\alpha},U_{\beta}\in \mathcal{U}$ such that $U_{\alpha},U_{\beta}\ne \varnothing$, we have $U_{\alpha}\cap U_{\beta}$ is path-connected. Then X is simply connected.

Proof: Let $\mathcal{U}$ be an open cover of $X$ such that each $U\in \mathcal{U}$ is simply connected, where $U_{\alpha}\cap U_{\beta}$ is nonempty and path connected for all nonempty $U_{\alpha},U_{\beta}\in \mathcal{U}$ and let $p\in U_{\alpha}\cap U_{\beta}$. Now, let $$f_{p,p}:[0,1]\to X$$ be a loop based at $p$. Since $f_{p,p}$ is continuous, the collection $\{f_{p,p}^{-1}(U):U\in \mathcal{U}\}$ is an open cover of the compact metric space $[0,1]\subset \mathbb{R}$. Since $[0,1]$ is compact it admits a finite subcover $$\bigcup_{i=1}^n f_{p,p}^{-1}(U_i)\supseteq [0,1]$$ and since it is a compact metric space, by the Lebesgue number Lemma $\exists \ \delta >0$ such that $\forall \ [a,b] \subseteq [0,1]$ with $diam([a,b])<\delta \implies [a,b]\in f_{p,p}^{-1}(U_i)$ for some $i\in\{1,\dots,n\}$, so select an integer $m$ such that $\frac{1-0}{m}=\frac{1}{m}<\delta$, and partition $$[0,1]\to t_0=0<t_1<t_2<\dots <t_m=1$$ by $$t_i=i\frac{1}{m},\mbox{ for }1\le i\le m$$ so that $$diam([t_i,t_{i+1}])=\frac{i+1}{m}-\frac{i}{m}=\frac{1}{m}<\delta, \ \forall \ i$$ and extending the smaller of $m$ or $n$ so that $N=\max\{m,n\}$ where we have $$\bigcup_{i=1}^N f_{p,p}^{-1}(U_i)=\bigcup_{i=0}^{N-1} f_{p,p}^{-1}(U_i)\supseteq [0,1], \quad [0,1]\to t_0=0<t_1<t_2<\dots <t_N=1$$ and hence $[t_i,t_{i+1}]\in f_{p,p}^{-1}(U_i)\implies f_{p,p}([t_i,t_{i+1}])\in U_i$ for some
$U_i\in \{U_0,U_2,\dots,U_{N-1}\}\subseteq \mathcal{U}$, then reordering the $U_i's$ if necessary so that $$f_{p,p}([0,t_1])\in U_0, \ f_{p,p}([t_1,t_2])\in U_1, \ \dots , \ f_{p,p}([t_{N-1},t_N])\in U_{N-1}$$ we would also like to arrange so that each $f_{p,p}(t_i)\in U_{i}\cap U_{i+1}$, where for the last point we seek $f_{p,p}(t_N)=f_{p,p}(1)=p\in U_{N-1}\cap U_0$. This is possible simply by noting that if $f_{p,p}(t_i)\notin U_{i}\cap U_{i+1}$ then each of $f_{p,p}([t_{i-1},t_{i}]),f_{p,p}([t_i,t_{i+1}])\in U_{i}$ or $f_{p,p}([t_{i-1},t_{i}]),f_{p,p}([t_i,t_{i+1}])\in U_{i+1}$, so the interval $f_{p,p}([t_{i-1},t_{i+1}])\in U_i\implies [t_{i-1},t_{i+1}]\in f_{p,p}^{-1}(U_i)$ and so $\{f_{p,p}^{-1}(U_0),\dots,f_{p,p}^{-1}(U_{N-1})\}\setminus \{f_{p,p}^{-1}(U_i)\}$ is still a finite open cover for $[0,1]$,
or $f_{p,p}([t_{i-1},t_{i+1}])\in U_{i+1}\implies [t_{i-1},t_{i+1}]\in f_{p,p}^{-1}(U_{i+1})$ and so $\{f_{p,p}^{-1}(U_0),\dots,f_{p,p}^{-1}(U_{N-1})\}\setminus \{f_{p,p}^{-1}(U_{i+1})\}$ is still a finite open cover for $[0,1]$.
Also, we observe that if $f_{p,p}([t_{i-1},t_{i+1}])\in U_{i}$ or $f_{p,p}([t_{i-1},t_{i+1}])\in U_{i+1}$, the Lebesgue number lemma is still satisfied with the point $f_{p,p}(t_i)$ removed, giving a new partition $t_{1_1},t_{1_2},\dots, t_{1_{N-2}}$, and we may repeat this process until we have obtained a partition, and finite open cover $$[0,1]\to t_{k_1},t_{k_2},\dots, t_{k_{N-k}},\quad \bigcup_{j=1}^{N-1-k} f_{p,p}^{-1}(U_{k_j})$$ satisfying $f_{p,p}(t_{k_j})\in U_{k_j}\cap U_{k_{j+1}}$ for each end point $j\in \{1,2,\dots, N-k-1\}$. So, WLOG let us assume we have obtained this with our initial partition.
Next, defining $$f_i:=\iota_{U_{i}} \circ f_{p,p}|_{[t_i,t_{i+1}]}:[t_i,t_{i+1}]\to X, \quad \forall \ 0\le i\le N-1$$ which is continuous as the restriction of a continuous map in $U_{i}\subseteq X$, and continuous in $X$ by the Characteristic property of the subspace topology.
or taking $L_i$ to be the linear map $L_i:[0,1]\to [t_i,t_{i+1}]$ $$f_i:=f_{p,p}\circ L_i:[0,1]\to X,\mbox{ by }f_i(s)=f_{p,p}\big(t_i+s(t_{i+1}-t_i)\big)$$ And note that for each $i$ we have $f_i(1)=f_{i+1}(0)$ so that $f_i*f_{i+1}$ is defined for each $i$. And therefore $$[f_{p,p}]=[f_0]*[f_1]*\cdots*[f_{N-1}]=\begin{cases}f_0(s), &s\in[0,t_1]\\ f_1(s), &s\in [t_1,t_2]\\ \quad \vdots &\qquad \vdots \\ f_{N-1}(s), &s\in [t_{N-1},t_N]\end{cases}$$ Which is also continuous by the Gluing Lemma for continuous maps.
Now, by construction we have arranged for $f_{p,p}(t_i)\in U_i\cap U_{i+1}\subset X, \ \forall \ 0\le i\le N$, and since $U_i\cap U_{i+1}$ is path-connected for each $i$, we may construct a path $\gamma_i:[0,1]\to U_i\cap U_{i+1}$ connecting $p$ to $f_{p,p}(t_i)$ entirely contained in $U_i\cap U_{i+1}$.
Note that $f_{p,p}(t_0)=p=f_{p,p}(t_N)$ and so we may select $\gamma_0=c_p=\gamma_N$ and so for each $0\le i\le N-1$ we have that $\gamma_{i}(1)=f_i(0)$ and $f_i(1)=\gamma_{i+1}^{-1}(0)$, so the products $\gamma_{i}*f_i$ and $f_i*\gamma_{i+1}^{-1}$ are defined. Then defining $$g_i:=\gamma_{i}*f_i*\gamma_{i+1}^{-1}$$ we have for each $i$ that $g_i\in \pi_1(U_{i},p)$ such that $U_{i}\in\{U_0,\dots,U_{N-1}\}$, depending on $f_i$. And since each $U_i$ for $0\le i\le N-1$ is simply connected we have for each $i$, that $[g_i]\sim [c_p]$. And so we have \begin{align*} [c_p]&=[c_p]*[c_p]*\dots*[c_p]\\ &\sim [g_0]*[g_1]*\dots *[g_{N-1}]\\ &=[\gamma_{0}*f_0*\gamma_1^{-1}]*[\gamma_{1}*f_1*\gamma_2^{-1}]*\dots*[\gamma_{N-1}*f_{N-1}*\gamma_{N}^{-1}]\\ &=[\gamma_{0}*f_0*\gamma_1^{-1}*\gamma_{1}*f_1*\gamma_2^{-1}*\dots *\gamma_{N-1}*f_{N-1}*\gamma_{N}^{-1}]\\ &=[\gamma_{0}]*[f_0]*[\gamma_1^{-1}*\gamma_{1}]*[f_1]*\dots *[\gamma_{N-1}^{-1}*\gamma_{N-1}]*[f_{N-1}]*[\gamma_{N}^{-1}]\\ &=[c_p]*[f_0]*[c_p]*[f_1]*[c_p]*\dots *[c_p]*[f_{N-1}]*[c_p]\\ &=[f_0]*[f_1]*\cdots*[f_{N-1}]\\ &=[f_{p,p}] \end{align*} and so $[f_{p,p}]\sim [c_p]$ where $[f_{p,p}]\in \pi_1(U_{\alpha}\cap U_{\beta},p)$.
Since $p\in U_{\alpha}\cap U_{\beta}$ was arbitrary, as was $f_{p,p}\in \pi_1(U_{\alpha}\cap U_{\beta},p)$, we conclude that for each loop $f_{p,p}\in \pi_1(U_{\alpha}\cap U_{\beta},p)$ based at $p$ we have $[f_{p,p}]\sim [c_p]$.
Then as $U_{\alpha},U_{\beta}\in \mathcal{U}$ were arbitrary, we conclude that $\forall \ U_{\alpha},U_{\beta}\in \mathcal{U}$ such that $p\in U_{\alpha}\cap U_{\beta}$ we have $[f_{p,p}]\sim [c_p]$. That is $[f_{p,p}]\sim [c_p]$ where $[f_{p,p}]\in \pi_1(X,p)$ and therefore, $X$ is simply connected.