Let $A\in M_{n\times n}(\Bbb{R})$ and $B:\Bbb{R}^+\to M_{n\times n}(\Bbb{R})$ be continuous.
\begin{align}(a)\begin{cases}x'(t)=\left(A+B(t)\right)x(t) & t\geq 0,\\x(0)=x_0\in \Bbb{R}&\end{cases}\end{align}
We assume that $\sigma(A)\subset \{\lambda\in \Bbb{C}:\operatorname{Re}(\lambda)<0\}$ and $\lim\limits_{t\to\infty}\Vert B(t)\Vert=0.$
I want to prove that $x(t,x_0)\to 0$ as $t\to \infty$
MY TRIAL: The solution to $(a)$ is given by \begin{align} x(t,t_0)=e^{tA}x_0+\int^{t}_{0}e^{(t-s)A}B(s)x(s)ds\end{align}
\begin{align} \Vert x(t,t_0)\Vert &=\Vert e^{tA}x_0+\int^{t}_{0}e^{(t-s)A}B(s)x(s)ds\Vert \\&\leq \Vert e^{tA}x_0\Vert +\int^{t}_{0}\Vert e^{(t-s)A}\Vert \Vert B(s)\Vert \Vert x(s)\Vert ds \end{align}
Since $tA$ and $-sA$ commute, then $e^{(t-s)A}=e^{t A}e^{-s A}$. Applying sup norm, we get \begin{align} \Vert x(t,t_0)\Vert &\leq \Vert e^{tA}x_0\Vert +\Vert e^{t A}\Vert\int^{t}_{0}\Vert e^{-s A}\Vert \Vert B(s)\Vert \Vert x(s)\Vert ds \end{align} By Gronwall's Lemma, \begin{align} \Vert x(t,t_0)\Vert \leq \Vert e^{tA}x_0\Vert e^{\Vert e^{t A}\Vert\int^{t}_{0}\Vert e^{-s A}\Vert \Vert B(s)\Vert ds}\end{align} Since $\sigma(A)\subset \{\lambda\in \Bbb{C}:\operatorname{Re}(\lambda)<0\}$, then $\lim\limits_{t\to\infty}\Vert e^{tA}x_0\Vert=0.$
Honestly, I don't know where to go from here but I still have an unused assumption $\lim\limits_{t\to\infty}\Vert B(t)\Vert=0.$
Please, I need help to proceed! Thank you for your time! If I have gone the wrong way, kindly let me know too!
I think your approach should work, but I haven't been able to do it in an easy way... Here's an alternative approach: Lyapunov functions. Let $V(t)=\frac{1}{2}\|x(t)\|^2$, so that using the ODE gives $$ V'(t)=\langle x,x'\rangle=\langle x,Ax\rangle+\langle x,Bx\rangle. $$ If $-k$ is the largest eigenvalue of $A_s=\frac{1}{2}(A+A^t)$ (symmetric part of $A$), then $$ \langle x,Ax\rangle=\langle x,A_s x\rangle\le -k \|x\|^2=-2kV. $$
Choose now $T>t_0$ so that $B(t)$ is small, or more precisely, $\|B(t)\|<k/2$ for $t>T$. Then for large $t$: $$ V'(t)\le-2k V+kV=-kV,\qquad t>T. $$ From Gronwall, we get $V(t)\le e^{-k(t-T)}V(T)$. Since the ODE is linear, the solution cannot blow up, i.e. $V(T)<\infty$, and the stability result follows.
EDIT: it's easy enough to show $V(T)<\infty$. Let $C=\max_{t<T} \|B(t)\|$. Then as before $$ V'(t)\le -2kV+2CV=2(C-k)V,\qquad t<T $$ By Gronwall, $V(T)\le e^{2(C-k)T}V(0)$.