Hi,
Simply use Itô-Tanaka's formula I guess this should give something like : $df(S_t)=D_-f(S_t)dS_t+\frac{1}{2}dL^s_tf''(ds)$
with $f(S)=(S-K)^+$ so $D_-f(S)=1_{]K,+\infty}(S)$ and $f''(ds)=\delta_K(ds)$
This gives if I am not mistaken :
$d(S_t-K)^+=df(S_t)=1_{]K,+\infty}(S_t)dS_t+\frac{1}{2}dL^K_t$
With $L^K_t$ being the local time of your geometric Brownian Motion $S$ around level $K$ at time $t$.
Regards
Edit
NB:
-$D_-$ stands for the left derivative of $f$
-$f''(ds)$ stands for second derivatives in the distribution-sense.
-The use of Itô-Tanaka's formula allows to avoid the derivation of the Mollifiers-type argument for the direct proof of the result (which is quite cumbersome in my opinion). I should add that Ito-Tanaka's formula is applicable to every $f$ that is the differnce of two convex functions if I remember well, which is the case here with $f(x)=(X-K)^+$.
