## Yes we can!

There's possibly a much simpler proof. The argument below starts with the observation that if $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x(t)$ must be automatically sublinear. This can then be upgraded to show that $x(t)$ grows slower than any power of $t$ using Gronwall. And finally a finer analysis of the error term gives you boundedness. The key step is the Gronwall analysis. 

### Sublinear growth

First one observes that if the integral is finite, the growth rate is sublinear. This follows from the fact that if the integral $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x'(t) \to 0$ as $t\to\infty$, and hence $x(t) = o(t)$ by, e.g., L'Hopital. 

### Upgrade the growth control

Integrating the bound for $x'(t)$ gives, for arbitrary $1 \leq t_1 < t_2$

$$ x(t_2) - x(t_1) \leq \int_{t_1}^{t_2} \int_t^\infty x(s) k(s) s^{-2} ds~dt $$

Integrating in parts in $t$, using that $dt = d(t - t_1)$

$$ x(t_2) - x(t_1) \leq (t_2 - t_1) \int_{t_2}^\infty x(s) k(s) s^{-2} ds + \int_{t_1}^{t_2} (t - t_1) x(t) k(t) t^{-2} dt $$

Now, for any $\delta > 0$, there exists $\tau_{\delta}$ such that for every $s \geq \tau_{\delta}$, it holds that

$$ x(s) \leq \delta s, \qquad \text{and} \qquad \int_{s}^\infty k(s)s^{-1} ds < \delta $$

since we know that $x$ grows sublinearly and the relevant integral converges. This implies that, for $t_2 > \tau_{\delta}$, 

$$ x(t_2) \leq x(\tau_{\delta}) + (t_2 - \tau_{\delta}) \delta^2 + \int_{\tau_{\delta}}^{t_2} x(s) \cdot k(s) s^{-1} ds $$

By Gronwall's inequality this means that 

$$ x(t_2) \leq [ \delta \tau_{\delta} + (t_2 - \tau_{\delta}) \delta^2 ] e^{\delta} $$

This final inequality can be used to estimate $\tau_{\delta/2}$. Consider the inequality 

$$ [\delta \tau_{\delta} + (s - \tau_{\delta}) \delta^2] e^{\delta} \leq \frac{\delta}{2} s $$

Solving this inequality we see that this is satisfied whenever

$$   \frac{e^{\delta} \tau_{\delta}}{\frac12 - \delta e^{\delta}} \leq s $$

For convenience write 

$$ \frac{e^\delta}{\frac12 - \delta e^\delta} = 2^{1+\sigma(\delta)} $$

and note that $\sigma$ is continuous and $\sigma(0) = 0$. 

This implies that $\tau_{\delta/2} \leq 2^{1 + \sigma(\delta)} \tau_{\delta}$. 


Now, iterating this argument, starting from some sufficiently small $\delta$ that we fix, we see can construct a sequence of times $\tau_{2^{-k} \delta}$ such that for $T \geq \tau_{2^{-k} \delta}$ it holds

$$ x(T) \leq 2^{-k} \delta T $$

We also have by iteration the bounds

$$ \tau_{2^{-k}\delta} \leq 2^k 2^{\sum_{j = 0}^{k-1} \sigma(2^{-j} \delta)} \tau_{\delta} $$

Using that $\sigma$ is continuous and $\sigma(0) = 0$, this immediately implies that $x(t)$ must grow slower than any power of $t$. 

### Boundedness

To get boundedness, we need to control the sum 

$$ \sum_{j = 0}^{k-1} \sigma(2^{-j} \delta) $$

If this sum is uniformly bounded in $k$, then we have the uniform bound $\tau_{2^{-k}\delta} \leq C 2^{k} \tau_{\delta}$, and this will imply the uniform bound for $x$ of the form $x(T) \leq C \delta \tau_{\delta}$. 

Returning to the definition of $\sigma$ we see that 

$$ \sigma(\delta) = \frac{1}{\ln 2} ( \delta - \ln(1 - 2\delta e^{\delta}) ) $$

Thus for all sufficiently small $\delta$ we have

$$ \sigma(\delta) \leq \frac{1}{\ln 2} ( \delta + 2 \delta e^{\delta}) \leq \frac{4}{\ln 2} \delta $$

This implies the summability of $\sigma(2^{-j} \delta)$, and the desired conclusion.