By considering $M_n+10$ instead of $M_n$, we may and shall assume that $\{M_n\}$ is a Martingale with respect to $\{{\mathcal F}_n\}$ that satisfies $M_n \ge 0$ for all $n$.
We will show that
$$
\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{2}<\infty \quad \text { a.s. } \tag{1}
$$
which implies that the same holds when the exponent 2 is replaced by 4.
Given $L>E(M_0)$, let $\tau=\inf\{n \ge 0 :M_n \ge L \}$ where by convention, $\, \inf \, \emptyset=\infty$.
By optional stopping,
$$E(M_0)=E(M_{\tau \wedge n}) \ge P(\tau \le n)L \,,$$
so letting $n \to \infty$ gives
$$P(\tau <\infty) \le E(M_0)/L \,. \tag{2}$$
Next, write $M^{\tau}_n:=M_{\tau \wedge n}$ and $\delta_0:=E[M_0-(M_0 \wedge L)]$, and for $n \ge 1$,
$$\delta_n:=E[M^{\tau}_n-(M^{\tau}_n\wedge L)|{\mathcal F}_{n-1}] \ge 0\,.$$
Note that $\delta_n \le E[M^{\tau}_n |{\mathcal F}_{n-1}] \le L .$
Define a martingale $\{Y_n\}$ by $Y_0=M_0\wedge L$ and
$$Y_n=Y_{n-1}+(M^{\tau}_n \wedge L)-M^{\tau}_{n-1}+\delta_n$$
Induction on $n$ shows that
$$Y^{\tau}_n:=Y_{\tau \wedge n}=(M^{\tau}_n \wedge L)+\sum_{k=0}^{\tau\wedge n}
\delta_k \,. \tag{3}
$$
Clearly $Y^{\tau}_n \ge 0$ and $|Y^{\tau}_n-Y^{\tau}_{n-1}| \le 2L$ for all $n \ge 1$. Let
$$T=\tau \wedge \inf\{ n\ge 0 : Y^{\tau}_n \ge L\}\,,$$
so that $$0 \le Y^T_n=Y_{T \wedge n} \le 3L \tag{4}$$ for all $n$.
From $(3)$ we infer that
$$\sum_{k=0}^{T\wedge n}
\delta_k \le 3L \,. \tag{5}$$
Since $$E(M_0) \ge E(Y_0) =E(Y_{T\wedge n}) \ge L \cdot P(T \le n <\tau)$$
we deduce that $$P(T \le n) \le P(\tau \le n)+ P(T \le n <\tau) \le 2E(M_0)/L\,,$$ so
$P(T<\infty) \le 2E(M_0)/L$. By orthogonality of Martingale increments
$$9L^2 \ge E(Y_{T \wedge n}^2)=\sum_{k=1}^n E[(Y_{T\wedge k}-Y_{T\wedge (k-1)})^2] \,. \tag{6}$$
Thus on the event $\{T=\infty\}$, we have
$$ (M_{i}-M_{i-1} )^{2} =
(Y_i-Y_{i-1}-\delta_i)^2 \le 2(Y_i-Y_{i-1})^2 +2\delta_i^2 \le 2(Y_i-Y_{i-1})^2 +2L\delta_i \,$$
so
$$\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{2} \le
2\sum_{i=1}^{\infty} [(Y_i-Y_{i-1})^2 +L\delta_i] <\infty \; \text{a.s.}$$
by $(5)$ and $(6)$. Thus
$$P\Bigl(\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{2} =\infty\Bigr) \le P(T<\infty) \le 2E(M_0)/L\,.$$
Taking $L \to \infty$ shows this probability is $0$.
For any process $X=(X_n)_{n \ge 0}$, write
$$G_n(X):=\sum_{i=2}^{n}\left(X_{i-2}\cdot (X_{i}-X_{i-1})\right)\,. \tag{1}$$
If $\, \widetilde{M}_n:=M_n+10 \, \,$ for all $n$, then
$$G_n(\widetilde{M})-G_n(M)=10 \sum_{i=2}^{n} \left(\widetilde{M}_{i}-\widetilde{M}_{i-1} \right)=10(\widetilde{M}_n-\widetilde{M}_1)$$
which converges almost surely, since non-negative martngales converge by the Martingale convergence theorem. Therefore, to prove that $G_n(M)$ tends to a finite limit a.s., it suffices to prove that
$\lim_n G_n(\widetilde{M})<\infty\, \, $ a.s.
Thus it will be sufficient to prove the following
Claim
Given a Martingale $(M_n)$ with respect to $({\mathcal F}_n)$ such that $M_n \ge 0$ for all $n$, we have that $\lim_n G_n( M )<\infty \, \, $ a.s.
The proof below is not short, but the idea is simple: We truncate $M$ at a height $L$ that it is unlikely to reach, correct the truncation to be a bounded Martingale $Y^T$, prove that $G_n(Y^T)$ converges almost surely, and then handle the errors due to the correction.
Proof of claim:
We first recall the construction from Almost sure convergence of martingale increment , which is included here for convenience:
Given $L>E(M_0)$, let $\tau=\inf\{n \ge 0 :M_n \ge L \}$ where by convention, $\, \inf \, \emptyset=\infty$.
By optional stopping,
$$E(M_0)=E(M_{\tau \wedge n}) \ge P(\tau \le n)L \,,$$
so letting $n \to \infty$ gives
$$P(\tau <\infty) \le E(M_0)/L \,. \tag{2}$$
Write $M^{\tau}_n:=M_{\tau \wedge n}$ and $\delta_0:=E[M_0-(M_0 \wedge L)]$, and for $n \ge 1$,
$$\delta_n:=E[M^{\tau}_n-(M^{\tau}_n\wedge L)|{\mathcal F}_{n-1}] \ge 0\,.$$
Note that $\delta_n \le E[M^{\tau}_n |{\mathcal F}_{n-1}] \le L .$
Define a martingale $\{Y_n\}$ by $Y_0=M_0\wedge L$ and
$$Y_n=Y_{n-1}+(M^{\tau}_n \wedge L)-M^{\tau}_{n-1}+\delta_n$$
Induction on $n$ shows that
$$Y^{\tau}_n:=Y_{\tau \wedge n}=(M^{\tau}_n \wedge L)+\sum_{k=0}^{\tau\wedge n}
\delta_k \,. \tag{3}
$$
Clearly $Y^{\tau}_n \ge 0$ and $|Y^{\tau}_n-Y^{\tau}_{n-1}| \le 2L$ for all $n \ge 1$. Let
$$T=\tau \wedge \inf\{ n\ge 0 : Y^{\tau}_n \ge L\}\,,$$
so that $$0 \le Y^T_n=Y_{T \wedge n} \le 3L \tag{4}$$ for all $n$.
From $(3)$ we infer that
$$\Delta_n:=\sum_{k=0}^n \delta_k \quad \text{satisfy} \quad \Delta_{T\wedge n}
\le 3L \quad \text{for all } \; n\,. \tag{5}$$
Since $$E(M_0) \ge E(Y_0) =E(Y_{T\wedge n}) \ge L \cdot P(T \le n <\tau)$$
we deduce that $$P(T \le n) \le P(\tau \le n)+ P(T \le n <\tau) \le 2E(M_0)/L\,,$$ so
$$ P(T<\infty) \le 2E(M_0)/L \,. \tag{*} $$
The two processes $(Y^T_n)$ and $\bigl(G_n(Y^T)\bigr)$ are both martingales with respect to $({\mathcal F}_n)$. Moreover, by orthogonality of Martingale increments and $(4)$,
$$E\Bigl[ G_n^2(Y^T)\Bigr] =\sum_{k=2}^n E \Bigl[ \bigl(Y^T_{ k-2 }\bigr)^2 \bigl(Y^T_{k}-Y_{k-1}^T \bigr)^2 \Bigr] \le 9L^2\cdot \sum_{k=2}^n E\Bigl[ \bigl(Y^T_{k}-Y_{k-1}^T \bigr)^2 \Bigr] \le 9L^2 E(Y^2_{T \wedge n}) \,.$$
By applying $(4)$ again we deduce that $E\Bigl[ G_n^2(Y^T)\Bigr] \le 81L^4$, so
$$ \lim_n G_n(Y^T) < \infty \quad \text{a.s.} \,. \tag{6} $$
On the event $\{T=\infty\}$, we have $M_n=Y_n-\Delta_n$ for all $n$, so
$$M_{k-2}(M_k-M_{k-1})=(Y_{k-2}-\Delta_{k-2})\cdot(Y_k-Y_{k-1}-\delta_k) \,,$$
whence
$$G_N(M)=G_n(Y)-\sum_{k=2}^n (Y_{k-2}-\Delta_{k-2})\delta_k -\sum_{k=2}^n \Delta_{k-2} \cdot(Y_k-Y_{k-1})\,. \tag{7} $$
On the event $\{T=\infty\}$, we have for $m<n$ that
$$|\sum_{k=m}^n (Y_{k-2}-\Delta_{k-2})\delta_k | \le 3L \sum_{k=m}^n \delta_k$$
which tends to $0$ as $m,n \to \infty$ by $(5)$.
Since Cauchy sequences converge,
$$\lim_n \sum_{k=m}^n (Y_{k-2}-\Delta_{k-2})\delta_k <\infty \,. \tag{8}$$
Observe that $\sum_{k=2}^n \Delta_{(k\wedge T)-2} \cdot(Y_k^T-Y_{k-1}^T)$
is an $L^2$ bounded martingale, so it converges a.s. Therefore,
$\sum_{k=2}^n \Delta_{k-2} \cdot(Y_k-Y_{k-1})$ converges a.s. on the event
$\{T=\infty\}$,
Combining the last observation with $(6), (7), (8)$, we conclude that
$$P(\lim_n G_n(M) <\infty) \le P(T<\infty) \le 2E(M_0)/L $$
by $(*)$. Since $L$ can be chosen arbitrarily large, this proves the claim.
Best Answer
Since $$ \mathsf{E}[M_1\mid M_0]=M_0, $$ $$ \delta:= \mathsf{P}(M_1\ge 1\mid M_0=1)>0. $$ Now, for $n\ge 2$, \begin{align} \mathsf{P}(|M_n| > 0.1)&\ge \mathsf{P}(M_n>0.1) \\ &\ge \mathsf{P}(M_n>0.1, M_0=1,M_1\ge 1) \\ &=\mathsf{P}(M_n>0.1 \mid M_0=1,M_1\ge 1) \\ &\quad \times\mathsf{P}(M_0=1,M_1\ge 1) \\ &=1\times \delta/2 \not\to 0 \end{align} as $n\to\infty$.