Your intuition is good, but there are some technicalities you should be careful about. WLOG $M_0 = 0$. Start by assuming $M$ is a continuous martingale of bounded variation. Then if $B$ is a bound on the variation of $M$, and $(t_i)$ is a partition of $[0,t]$,
$$
E[M_t^2] = E\sum (M_{t_{i+1}}-M_{t_i})^2 \leq B E[\sup_i|M_{t_{i+1}}-M_{t_i}|].
$$
Since $M$ is continuous the supremeum tends to $0$ as the partition size goes to $0$. Moreover, the supremum is dominated by $B$, and hence dominated convergence tells us that $E[M_t^2] = 0$, in particular $M_t=0$ a.s. Let $t$ run over rationals and use the continuity of $M$ to conclude that $M=0$.
Next, if $M$ is a continuous local martingale, take a localizing sequence of stopping times $(\tau_n)$. We then find that $M^{\tau_n}_t = 0$ for all $n$, and so taking the limit in $n$ gives $M_t = 0$ a.s. Again let $t$ range over rationals.
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.
Best Answer
No, not every continuous $L^2$ bounded local martingale is a true martingale (see your counterexample). However, every local martingale with $L^1$ bounded quadratic variation is a true martingale!
The problem is in your second statement:
If $X$ is a continuous martingale with $L^2$-bounded quadratic variation: $\mathbb{E}[X]_\infty <\infty$ THEN $X_t$ is $L_2$ bounded. The converse ('if f ') is not true in general (which is what is going wrong in your example).
Let $\tau_n$ be a localizing sequence. we then have:
$$ \mathbb{E}X_t^2 = \mathbb{E}\lim_{n \rightarrow \infty} X_{t \wedge \tau_n}^2 \leq \lim_{n \rightarrow \infty} \mathbb{E} X_{t \wedge \tau_n}^2 = \lim_{n \rightarrow \infty} \mathbb{E} [X]_{t \wedge \tau_n}= \mathbb{E} \lim_{n \rightarrow \infty} [X]_{t \wedge \tau_n} =E[X]_t $$
The inequality follows from Fatous lemma. The second exchange of limit and integration is, however, justified by the monotone convergence theorem.
In your specific example, Fatou's lemma results in a strict inequality:
$$ \frac{1}{1+t} = \mathbb{E} Y_t^2 < \mathbb{E}[Y]_t = \mathbb{E} \int_0^t \frac{1}{X_{1+t}^4} ds = \int_0^t \mathbb{E}Y_s^4 ds = \infty $$ The exchange of integrators is justified because $Y_s^4$ is positive. Further, while the second moment of $Y_s$ exists, the fourth moment does not. Hence the last value is infinity.