Well, in the form stated above, none of the statements are true, because you're only assuming $f$ to be progressive and not predictable, and you're not assuming that the integrator $X$ has continuous sample paths.
I'd say that point (4) is neither true nor false but undefined, as the stochastic integral is not necessarily well-defined for integrands which are only progressive and not predictable.
As regards the other three points, as a counterexample, take e.g. $N$ to be a standard Poisson process and let $f(t) = N_t$, $X_t = N_t - t$ and let $(\mathcal{F}_t)$ be the filtration induced by $N$. Then $f$ is locally bounded (by e.g. the sequence of stopping times corresponding to the jump times of $N$), bounded on compacts (because it has cadlag sample paths) and is progressive (because it is cadlag and adapted). Furthermore, the integral is well-defined since $X$ has sample paths of finite variation, so the integral can be defined as a pathwise Lebesgue integral. It holds that
$$
Y_t = \int_0^t f(s) dX_s = \int_0^t (N_{s-} + \Delta N_s) dX_s \\
= \int_0^t N_{s-}dX_s + \sum_{0<s\le t}(\Delta N_s)^2
= \int_0^t N_{s-}dX_s + N_t.
$$
This functions as a counterexample for points (1-3) because even though $X$ is a locally $L^2$-bounded martingale, $Y$ is not even a local martingale. The problem is that $f$ is not predictable. See also this question for more on this.
If $f$ was assumed predictable, the answers would be:
(1): True. Intuitively, this is because the integral of a predictable process with respect to a local martingale is a martingale, and if $f$ is sufficiently rough, the integral process will not yield integrability, and so a true martingale cannot be expected.
(2): True. Intuitively, this is because the integral process is a local martingale, and by localising so that $f$ is bounded and $X$ is $L^2$-bounded, one obtains $L^2$ boundedness of the integral process.
(3): True. This is almost a defining property of the stochastic integral (depending on the method of construction), but certainly true in any case.
(4): True, also almost by construction, depending on the method of construction.
If a progressively measurable $(H_s)_{s\in[0,t]}$ satisfies $\int_0^t\vert H_s\vert^2\,d\langle M\rangle_s<+\infty$ almost surely, then the stochastic integral is a local martingale. If moreover $\mathbb E\left[\int_0^tH_s^2\,d\langle M\rangle_s\right]<+\infty$, then it is a martingale.
Here you would need that $\mathbb E\left[\int_0^tM_s^{4p-2}\,d\langle M\rangle_s\right]$ to conclude that it is a martingale.
Best Answer
Given any $a>0$ any local martingale can be decomposed into a local martingale which is of finite variation and a local martingale (starting at zero) with jumps bounded by $a$.
A local martingale $M$ whose jumps are bounded and with $M_0 \in L^2$ belongs to $H^2_{loc}$.
For the part $M$ with finite variation you can use the fact that if $H$ is a locally bounded predictable process, then the compensator $(H \cdot M)^p$ of the process $H \cdot M$ is $H \cdot M^p$ (and $M^p$- i.e the compensator of $M$ is obviously zero).