For completeness, I will include the proof from Revuz, Yor "Continuous martingales and Brownian Motion" here.
$M-[M]$ is a local martingale, so there exists ${T_n},n\ge1$ such that $T_n\to\infty$ and $\mathbb{E}[M^2_{t\wedge T_n}]=\mathbb{E}([M]_{t\wedge T_n})<K$. Apply Fatou: $\mathbb{E}[M^2_t]\leq \liminf\mathbb{E}[[M]_{t\wedge T_n}]\leq K$. So $M$ is $L^2$-bounded.
Need to show now that $M$ is a true martingale. Let $T^{\prime}_n$ be stopping times reducing $M$, then $\{M_{t\wedge T^{\prime}_n}\}$ is UI, since it is $L^2$-bounded and $M_{t\wedge T^{\prime}_n} \to M_t$ in $L^1$, and so $\mathbb{E}[M_{t}|\mathcal{F_s}]=M_s$ a.s.
Edit
The omitted steps are:
$\mathbb{E}[M_{t\wedge T^{\prime}_n}|\mathcal{F}_s]=M_{s\wedge T^{\prime}_n}$, so
$\mathbb{E}[M_{t\wedge T^{\prime}_n}\mathbb{1}_A]=\mathbb{E}[M_{s\wedge T^{\prime}_n}\mathbb{1}_A]$ for all $A \in \mathcal{F}_s$. Now take $n\to \infty$, to conclude $\mathbb{E}[M_{t}\mathbb{1}_A]=\mathbb{E}[M_{s}\mathbb{1}_A]$ for all $A \in \mathcal{F}_s$, hence $\mathbb{E}[M_t|\mathcal{F}_s]=M_s$ a.s.
Let $(\sigma_n)_{n \in \mathbb{N}}$ be a localizing sequence of the local martingale $M$, i.e. an increasing sequence of stopping times such that $\sigma_n \to \infty$ and $(M_{t \wedge \sigma_n})_{t \geq 0}$ is a martingale for each $n \in \mathbb{N}$.
Applying the optional stopping theorem, we find that $(M_{t \wedge \sigma_n \wedge T(s)})_{t \geq 0}$ is a martingale, i.e.
$$\mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = M_{u \wedge \sigma_n \wedge T(s)} \tag{1}$$
for all $u \leq t$. Since $M$ has continuous sample paths, the right-hand side converges to $M_{u \wedge T(s)}$ as $n \to \infty$. If we can show that
$$\lim_{n \to \infty} \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = \mathbb{E}(M_{t \wedge T(s)} \mid \mathcal{F}_u)\tag{2}$$
this proves that $(M_{t \wedge T(s)})_{t \geq 0}$ is a martingale. To prove $(2)$, we note that by Doob's maximal inequality
$$\begin{align*} \mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge \sigma_n \wedge T(s)}|^2 \right) &\leq 4 \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)}^2) = 4 \mathbb{E} \big( [M]_{t \wedge \sigma_n \wedge T(s)} \big) \leq \mathbb{E}([M]_{T(s)}) \leq 4s. \end{align*}$$
By the monotone convergence theorem this implies
$$\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right) \leq 4s < \infty.$$
Hence,
$$\mathbb{E} \left( \sup_{n \in \mathbb{N}} |M_{t \wedge \sigma_n \wedge T(s)}| \right) \leq \sqrt{\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right)} < \infty.$$
Now $(2)$ follows from the dominated convergence theorem.
Best Answer
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.