= Solution
The usual <quadratic variation> convention includes $\langle M\rangle_0=0$. If $A,C$ are two continuous <finite-variation processes> with $A_0=C_0=0$ such that $M^2-A$ and $M^2-C$ are <continuous local martingales>, then $A-C$ is a <continuous local martingale> of <finite variation>. It must be constant, and its initial value is zero, so $A=C$ up to <indistinguishability of stochastic processes>.
Here is a proof of the fact used. Subtract the initial value of a <continuous local martingale> $N$ of <finite variation>, and stop so that both $|N|$ and its <total-variation process> are bounded by deterministic constants. The stopped <local martingale> is a bounded true <martingale>, so orthogonality of its increments gives, on any deterministic <partition of an interval> of $[0,t]$,
$$
\mathbb E(N_t^2)=\mathbb E\sum_i(\Delta_iN)^2\leq\mathbb E\left[\max_i|\Delta_iN|\,V_t(N)\right]\longrightarrow0.
$$
Continuity and the <dominated convergence theorem> justify the last limit. Remove stopping and use continuity on a countable dense set of times to conclude $N\equiv0$.
\b[As literally printed, uniqueness is false without normalization.] If $A$ works then $A+c$ also works for any deterministic constant $c$, since subtracting a constant preserves the <local martingale> property; this is a genuine omitted convention.
\b[With normalization, existence always holds for a <continuous local martingale>.] Its <quadratic variation> $[M]$, obtained as the limit in <probability> of squared increments, is continuous, increasing, starts at zero and satisfies $M^2-[M]$ is a <continuous local martingale>, so taking $\langle M\rangle=[M]$ supplies the required <finite-variation process>.
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