If any locally constant sheaf on $X$ is constant, then $X$ is simply connected

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I am having some trouble completing the proof of the following statement:

Let $X$ be a locally arcwise connected, connected topological space and suppose there exists an open covering stable by finite intersections by connected and simply connected subsets. If any locally constant sheaf on $X$ is constant, then $X$ is simply connected.

What I think I have understood from the proof of the notes I am following ($\textit{Algebra and Topology}$ by Pierre Schapira) is the next.

Firstly, we assume X is non empty and choose $x\in X$. Since $X$ is connected and locally arwise connected, it is arcwise connected and then the groupoid $\Pi_1(X)$ is equivalent to the category $\Pi_1(X,x)$ associated with the group $ \pi_1(X,x)$. That is, the category with one object $\{x\}$ and morphisms $\pi_1(X,x)$. Thus, the category $Fct(\Pi_1(X), \text{Mod}(k))$ is equivalent to $Fct(\Pi_1(X,x),\text{Mod}(k))$, and then to the category of representations $\text{Rep}(\pi_1(X,x), \text{Mod}(k))$, where $k$ denotes a commutative unitary ring. We also have that $\text{Mod}(k)$ is equivalent to $\text{Rep}(\{1\}, \text{Mod}(k))$ of trivial representations.

We are assuming that any locally constant sheaf on $X$ is constant. Therefore, we have the equality LCSH$(k_X)=$ CSH$(k_X)$. It is known that, if $X$ is non empty and connected, the category of constant sheaves CSH$(k_X)$ is equivalent to $\text{Mod}(k)$. From a previous theorem, thanks to the hypothesis made on $X$, LCSH$(k_X)$ is equivalent to $Fct(\Pi_1(X), \text{Mod}(k))$. Thus, we get an equivalence between the categories $Fct(\Pi_1(X), \text{Mod}(k))$ and $\text{Mod}(k)$.

Following all the equivalences, we find $\text{Rep}(\pi_1(X,x), \text{Mod}(k))$ is equivalent to $\text{Rep}(\{1\}, \text{Mod}(k))$.

From this I need to deduce that $\pi_1(X,x)$ is the trivial group with one element. I have tried to use the equivalence between them but I don't get anything. In addition, I don't know if all the previous arguments are right.