= Solution
For $g(x)=\log|x|$ on $\mathbb R^2\setminus\{0\}$,
$$
\nabla g(x)=\frac{x}{|x|^2},\qquad \Delta g(x)=0.
$$
The <Itô formula>, stopped in a compact <annulus>, therefore gives
$$
dM_t=\frac{B_t}{|B_t|^2}\mathbin\cdot dB_t.
$$
This proves the <local martingale> property up to a possible hitting of the origin. For the <annulus> exit time $T$, $M_{t\wedge T}$ is bounded between $\log r$ and $\log R$, so it is a true <uniformly integrable martingale>.
The exit is finite almost surely. Indeed $|B_t|^2-2t$ is a <martingale>, so stopping at $T\wedge t$ gives $2\mathbb E(T\wedge t)=\mathbb E|B_{T\wedge t}|^2-1\le R^2-1$. Letting $t\to\infty$ proves $\mathbb ET<\infty$. Bounded convergence in the logarithmic <martingale> now yields
$$
\boxed{\mathbb E M_T=M_0=0.}
$$
If $p$ is the <probability> of reaching radius $r$ before radius $R$, <continuity> gives
$$
p\log r+(1-p)\log R=0,\qquad \boxed{p=\frac{\log R}{\log(R/r)}}.
$$
Hitting zero before reaching radius $R$ would require reaching every positive radius $r$ first. Its <probability> is bounded by $p$, which tends to zero as $r\downarrow0$. Any finite-time path hitting zero is bounded before that time and hence hits zero before exiting some integer-radius disk. A countable union over those disks proves \b[$B_t\ne0$ for every $t\ge0$, almost surely]. This is <planar Brownian motion avoids a fixed point>; it also makes $M$ a globally defined <continuous local martingale>.
To decide whether it is a true <martingale>, compute its <expectation> directly. Condition on $B_0$ if necessary and use rotation to take $B_0=(1,0)$ and write $B_t=B_0+W_t$. Conditional on $\rho=|W_t|$, the angle of $W_t$ is uniform. The <angular average of a logarithmic potential> is
$$
\frac1{2\pi}\int_0^{2\pi}\log|1+\rho e^{i\theta}|d\theta=\log\max(1,\rho).
$$
For $\rho<1$ this follows by taking the real part of the convergent <power series> for $\log(1+\rho e^{i\theta})$: each nonconstant Fourier term integrates to zero. For $\rho>1$ factor out $\rho$ and apply the same argument to $1/\rho$. The equality at one follows by the <integrable> limiting logarithmic singularity. All interchanges are valid: the shifted <Gaussian density> is bounded near the origin, where $\int_0^1r|\log r|dr<\infty$, and its <Gaussian> tail controls logarithmic growth.
The radius $\rho$ has density $(\rho/t)e^{-\rho^2/(2t)}$. Consequently, for $t>0$,
$$
\boxed{\mathbb E M_t=\int_1^\infty\log\rho\,\frac{\rho}{t}e^{-\rho^2/(2t)}d\rho=\int_1^\infty\frac{e^{-\rho^2/(2t)}}{\rho}d\rho>0.}
$$
Since $M_0=0$, \b[$M$ is not a <martingale>], despite <integrability> at each deterministic time. This is the <logarithmic radius of planar Brownian motion> example.
The optional hint is consistent with this conclusion. The upper-radius exit $T(2,0)$ is finite and never occurs at zero. After that exit the stopped square is $(\log2)^2$; before it, a radius in $[1/2,2]$ also gives a logarithmic square no larger than $(\log2)^2$, while a radius below $1/2$ contributes precisely the extra indicated term. This proves the hinted inequality without assuming that the unstopped process is a true <martingale>.
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