Help with the determination of the interval of solutions to a Cauchy problem

Question

I am studying Differential Equations and I'm new to Cauchy problems and intervals determination. There is a problem that I am working on and I have some questions and I need a little bit of help.
Let $a: [0,+\infty[\rightarrow\mathbb{R}$ be a continuous function such that $|a(t)|\leq1$ for all $t\geq0$. This is my Cauchy problem:
\begin{align} (E):\bigg\{\begin{split} &x'=-x+a(t)x^2 \\ &x(0)=x_0 \end{split} \end{align} with $0<x_0<1$ .
I showed the existence and the uniqueness of the maximal solution in an interval $]T^-,T^+[$. Using that, I showed that $x(t)>0$ for all $t\in]T^-,T^+[$. I want to show now $x(t)\leq1$ for all $t\in]T^-,T^+[$. They supposed the opposite and considered this set: $$S=\{t\in[0,T^+[ \text{ such that } x(s)\leq1,\forall s\in[0,t]\}$$ I showed that $S$ is nonempty and that $T<T^+$ with $T=supS$. Now, I need to show that $x(T)=1$.
My try goes as following:
$\exists t_0\in[0,T^+[ \text{ s.t. } x(t_0)>1$ and since $x(t)$ is continuous then $\exists t_1\in[0,T^+[$ s.t. $x(t_1)=1$. Therefore, $t_1\in[0,T]$ i.e. $t_1\leq T$. And we know that $x_0<1$.
I am stuck here. I feel like I am really close to the answer but I just can't quite put my fingers on it. Besides, let's suppose that we proved that. I showed that for all $t\in[0,T],\space x'(t)\leq0$ thus $x(T)<1$. From that I will conclude that $x(t)\leq1$. Now, they want me to determine $T^-$ and $T^+$ and I don't know what kind of tricks I need to use to show that. It looks easy, not gonna lie.