= Solution
For $x\leq0$, $f_n(x)=|x|$; for $x>0$, only the interval $(0,1/n)$ contributes an error to the integral of the derivative. Hence
$$
0\leq|x|-f_n(x)\leq\frac2n\qquad(x\in\mathbb R).
$$
By the <Itô formula>, the continuous processes
$$
C_t^{(n)}=\frac12\int_0^t f_n''(X_s)d[M]_s=f_n(X_t)-\int_0^t f_n'(X_s)dM_s-\int_0^t f_n'(X_s)dA_s
$$
are nondecreasing, since $f_n''\geq0$. Parts (a) and (b), together with the uniform approximation of absolute value, show that $C^{(n)}$ converges <uniformly on compacts in probability> to
$$
Z_t=|X_t|-\int_0^t\operatorname{sgn}_-(X_s)dM_s-\int_0^t\operatorname{sgn}_-(X_s)dA_s.
$$
The limit is continuous: the <martingale> integral is continuous, and the bounded-integrand finite-variation integral is continuous because $A$ is continuous. For any rational $s<t$, <convergence in probability> and $C_t^{(n)}-C_s^{(n)}\geq0$ imply $Z_t-Z_s\geq0$ almost surely. Intersect these countably many events and use continuity to extend to all real $s<t$. Thus
$$
\boxed{Z_0=0,\qquad Z\text{ is continuous and nondecreasing almost surely}.}
$$
Since $|X|=\operatorname{sgn}_-(X)\cdot M+(\operatorname{sgn}_-(X)\cdot A+Z)$ is a <continuous local martingale> plus continuous <finite variation>, $|X|$ is a <continuous semimartingale>. This is the <smooth convex approximation proof of the Tanaka formula>. With the specified sign value $-1$ at zero, $Z$ is the <right local time> at zero, rather than an unspecified symmetric normalization.
For an arbitrary <continuous semimartingale> starting at zero, use its continuous decomposition $X=M+A$, with both parts starting at zero, and let
$$
\rho_k=\inf\{t\geq0:|M_t|+V_t\geq k\}\wedge k.
$$
The continuity and local finiteness of variation imply $\rho_k\uparrow\infty$ almost surely. Up to $\rho_k$ the required deterministic bound holds, and the stopped <local martingale> is bounded, hence a true $L^2$-bounded <martingale>. Apply the previous argument to $X^{\rho_k}$ for each integer $k$. Stopping the defining integrals gives precisely $Z_{t\wedge\rho_k}$. On the intersection of their full-probability events, each stopped $Z$ is nondecreasing; every finite interval is eventually contained before $\rho_k$, so $Z$ is nondecreasing globally. Therefore
$$
\boxed{|X|\text{ is a continuous semimartingale for every continuous semimartingale }X.}
$$
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