If a system is described by an impulse response $h(t)$, then by definition it is an LTI system. Because impulse response of the form $h(t) = \mathcal{T} \{ \delta(t) \}$ only exists for LTI systems.

$h(t)$$\mathrm{rect}(t-0.5)$

$$y(t) = \int_{0}^{\infty} x(\tau)\mathrm{rect}(t-0.5-\tau) \ \mathrm{d}\tau$$

Note that if $t<0$$t-0.5-\tau<-0.5$$\tau\in(0,\infty$

$$|t-0.5-\tau|>|0.5|$$

This means that the $\mathrm{rect}()$$t$$y(t)=0 \ \forall t<0$

For $t>0$$x(t)u(t)$$0$$-\infty$

If we call the window $w(t)$$y(t)$

$$y(t)= \left\{ \begin{array}{ll} [x(t)u(t)]*w(t) & \mbox{if } t > 0 \\ 0 & \mbox{if } t < 0 \end{array} \right.$$

Now, could you determine whether the system is time-invariant or not?

Hint: try appling a delay to the input signal, and see if its output is equal to a delayed version of the output itself.