From scaling relations, we may assume that $T=M=1$.

**** The argument I sketch below only gives a weaker result than in the OP, so the question is still open. Thanks to Nina for pointing this out. ****

Step 1: Introduce the random variable
$$Z_\epsilon=\sup_x \int_0^\epsilon 1_{|B_t-x|<2\epsilon} dt
\leq \sup_{j} \int_0^\epsilon 1_{|B_t-j\epsilon|<4\epsilon}dt, $$
and set $M_\epsilon=Z_\epsilon/\epsilon^{3/2}$. Using the heat kernel and the second inequality above, we obtain that $EM_\epsilon\leq C_1$ and
$EM_\epsilon^2\leq C_2$. (This requires a detailed computation, which I am skipping here, so this needs to be double checked. There may be a log correction.)

Step 2: Let $M_\epsilon^{(i)}$ be independent copies of $M_\epsilon$. By conditioning on the Brownian motion at times $(i-1)\epsilon$ and forgetting about the Lipschitz condition between the intervals, you get that the random variable in the OP is stochastically dominated by
$$\epsilon^{1/2} \cdot \epsilon \sum_{i=1}^{1/\epsilon} M_\epsilon^{(i)}=: \epsilon^{1/2} S_{\epsilon}\,.$$

Step 3: We have that $ES_\epsilon=C_1$ and therefore, by Markov's inequality and the estimate on $EM_\epsilon^2$,
$$P(S_\epsilon>2C_1)\leq P(S_\epsilon-ES_\epsilon\geq C_1)\leq \epsilon \frac{C_2}{C_1^2}\,.$$

Step 4: By interpolation, it is enough to consider the sequence $\epsilon_j=2^{-j}$ and apply Borel-Cantelli to conclude the estimate $\epsilon^{1/2}$ in the right hand side of the OP.