= Solution
Work first with $M_0=0$. More generally, the assertion needs $M_0\in L^2$ and then applies to $M-M_0$. Without this assumption, an integrable but non-square-integrable initial random variable kept constant in time has zero <quadratic variation> but is not bounded in <L2 space>.
Choose increasing <stopping times>
$$
\tau_n=\inf\{t:|M_t|\geq n\text{ or }[M]_t\geq n\}.
$$
They tend to infinity almost surely, because the continuous path and its <quadratic variation> are finite on every compact time interval. The stopped process is a bounded <martingale>, and $M_{t\wedge\tau_n}^2-[M]_{t\wedge\tau_n}$ is bounded and a <local martingale>. Taking expectations gives
$$
\mathbb E M_{t\wedge\tau_n}^2=\mathbb E[M]_{t\wedge\tau_n}.
$$
Suppose $C=\mathbb E[M]_\infty<\infty$. These squared expectations are at most $C$. For each fixed $t$, the stopped values converge almost surely to $M_t$ and are uniformly integrable, since they are uniformly bounded in <L2 space>. They therefore converge in $L^1$. Passing in the stopped conditional-expectation identity, using contraction of <conditional expectation> in $L^1$, gives $\mathbb E(M_t\mid\mathcal F_s)=M_s$ for $s\leq t$. Thus $M$ is a true <martingale>. The <Fatou lemma> gives $\mathbb E M_t^2\leq C$, uniformly in $t$, so it is an <L2-bounded continuous martingale>.
Conversely, suppose $M$ is a true <martingale> with $\sup_t\mathbb E M_t^2\leq C$. For fixed $t$, bounded-time <optional sampling theorem> gives $M_{t\wedge\tau_n}=\mathbb E(M_t\mid\mathcal F_{t\wedge\tau_n})$. Conditional <Jensen inequality> then yields
$$
\mathbb E[M]_{t\wedge\tau_n}=\mathbb E M_{t\wedge\tau_n}^2\leq\mathbb E M_t^2\leq C.
$$
The brackets increase to $[M]_t$. Applying the <monotone convergence theorem> first in $n$ and then in $t\uparrow\infty$ proves $\mathbb E[M]_\infty\leq C<\infty$.
The exact identities and terminal limits can also be justified. On each fixed horizon $t$, the <Doob L2 maximal inequality> gives
$$
\mathbb E\sup_{s\leq t}|M_s|^2\leq4\mathbb E|M_t|^2<\infty.
$$
This is an integrable dominating variable for $M_{t\wedge\tau_n}^2$, so the <dominated convergence theorem> and increasing convergence of the brackets remove the stopping:
$$
\mathbb E M_t^2=\mathbb E[M]_t.
$$
The <L2 martingale convergence theorem> supplies $M_\infty$ with $M_t\to M_\infty$ almost surely and in <L2 space>. Consequently the squares converge in $L^1$, while $[M]_t\uparrow[M]_\infty$. We obtain the <integrable terminal bracket criterion> with the precise terminal identity
$$
\boxed{M\text{ is an }L^2\text{-bounded martingale}\iff\mathbb E[M]_\infty<\infty,\qquad\mathbb E M_\infty^2=\mathbb E[M]_\infty.}
$$
For $M_0\in L^2$ rather than zero, the final identity becomes $\mathbb E M_\infty^2=\mathbb E M_0^2+\mathbb E[M]_\infty$.
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