$\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\F}{\mathcal{F}}\newcommand{\Om}{\Omega}\newcommand{\Z}{\mathbb{Z}}$A counterexample can be obtained as follows. Let $T_1,T_2,\dots$ be independent (say) geometrically distributed random variables with fast growing means, say with $ET_i=2^i$. 

Let $Y_n=0$ for $0\le n\le T_1$. For time moments $n$ after that, let the values of $Y_n$ coincide with the positions of a simple random walk $W^{(1)}_\cdot$ starting from $0$ at time $T_1$ -- but only till the time, say $\nu_1$, of the first return of $W^{(1)}_\cdot$ to $0$. This is the first "step". 

After this, let $Y_\cdot$ stay at $0$ for time $T_2$. After that, let the values of $Y_n$ coincide with the positions of a simple random walk $W^{(2)}_\cdot$ starting in state $0$ at time $\nu_1+T_2$ -- but only till the time, say $\nu_2$, of the first return of $W^{(2)}_\cdot$ to $0$, where $W^{(2)}_\cdot$ is independent of $W^{(1)}_\cdot$. This is the second "step". 

Continue doing such "steps" indefinitely (assuming that the walks are independent of the $T_j$'s). Thus, we get a martingale $(Y_n)$, with respect to a certain filtration of $\sigma$-algebras. The walks occur increasingly rarely; they start at very uncertain times (with standard deviations $\asymp2^i$ and corresponding very flat distributions); and the walks last comparatively short times. Therefore, for any given large time moment $n$, $P(Y_n\ne0)$ is small. So, $Y_n\to0$ in probability. 

However, because the walks occur infinitely many times, clearly $Y_n\not\to0$ almost surely. 

--- 

Here are formal details. 

**Construction:** Let $T_1,T_2,\dots$ be independent geometrically distributed random variables (r.v.'s) (defined on some measurable space $(\Om,\F)$) with means $ET_i=1/p_i$, where $0<p_i<1/2$ and 
\begin{equation*}
	\sum_{i\in\N}p_i^{1/2}<\infty. \tag{1}
\end{equation*}
So, 
\begin{equation*}
	P(T_i=t)=p_i q_i^{t-1}\,1(t\in\N) \tag{2}
\end{equation*}
for real $t$, where $q_i:=1-p_i$. 
  
Let $R_1,R_2,\dots$ be independent Rademacher r.v.'s (defined on the same measurable space $(\Om,\F)$) that are independent of the $T_i$'s; so, $P(R_i=1|(T_j))=1/2=P(R_i=-1|(T_j))$ for all $i\in\N$. 

Define (the starting times of the walks) $S_1,S_2,\dots$ and (the finishing times of the walks) $F_0,F_1,F_2,\dots$ by the following recursion: 
\begin{equation*}
	F_0:=0
\end{equation*}
and, for $k\in\N$, 
\begin{equation*}
	S_k:=F_{k-1}+T_k,\quad F_k:=\inf\{n\in\N\colon n>S_k,W^{(k)}_n=0\}, \tag{3}
\end{equation*}
where 
\begin{equation*}
	W^{(k)}_n:=\sum_{i\in\N}R_i\,1(S_k<i\le n). \tag{3a}
\end{equation*}
Note that $F_k=\min\{n\in\N\colon n>S_k,W^{(k)}_n=0\}<\infty$ for all $k\in\N_0:=\{0\}\cup\N$ almost surely (a.s.), because the simple random walk is recurrent. So, $S_k<\infty$ for all $k\in\N$ a.s. 

Now, for all $n\in\N$ let 
\begin{equation*}
	Y_n:=\sum_{k\in\N}1(S_k<n\le F_k)W^{(k)}_n. \tag{4}
\end{equation*}
Note that a.s. at most one summand in (4) is nonzero, since $S_k=F_{k-1}+T_k>F_{k-1}$ a.s. for all $k\in\N$. 

To complete the construction, for each $n\in\N$ let 
\begin{equation*}
	\F_n:=\si(R_1,\dots,R_n,\{1(T_k\le j)\colon k\in\N,j\in\N,j\le n\}), 
\end{equation*}
the $\si$-algebra generated by $R_1,\dots,R_n,\{1(T_k\le j)\colon k\in\N,j\in\N,j\le n\}$. Clearly, $(\F_n)_{n\in\N}$ is a filtration. 

**Showing that $((Y_n,\F_n))_{n\in\N}$ is a martingale:**
Abusing notation as is commonly done, for a r.v. $X$ let us write $X\in\F_n$ to mean that $X$ is $\F$-measurable. Then obviously $1(F_0\le n)=1(0\le n)\in\F_0:=\{\Om,\emptyset\}\subseteq\F_1$ for all $n\in\N$. Using now the relations 
\begin{equation*}
	1(S_k\le n)=\sum_{j=1}^{n-1}1(T_k=j)1(F_{k-1}\le n-j),
\end{equation*}
$1(T_k=j)=1(T_k\le j)-1(T_k\le j-1)$, and 
\begin{equation*}
	\{F_k\not\le n\}=
\bigcap_{m=1}^n\Big(\{S_k\not\le m-1\}\cup\Big\{\sum_{i=1}^m R_i\,1(S_k\le i-1)\ne0\Big\}\Big)
\end{equation*}
for natural $k,j$ (which follow by (3) and (3a)), we conclude by induction on $k$ that 
\begin{equation*}
	\{S_k\le n\}\in\F_n,\quad \{F_k\le n\}\in\F_n \tag{5}
\end{equation*}
for all natural $k,n$. 

Hence, by (4) and (3a), the sequence $(Y_n)_{n\in\N}$ is adapted to the filtration $(\F_n)_{n\in\N}$. Moreover, by (4), for all $n\in\N$ we have 
\begin{equation*}
Y_{n+1}-Y_n=\sum_{k\in\N}R_{n+1}\,1(S_k\le n,F_k\not\le n);	
\end{equation*}
also, $R_{n+1}$ is independent of $\F_n$. So, in view of (5), $((Y_n,\F_n))_{n\in\N}$ is a martingale. 

**Showing that $Y_n\to0$ in probability (as $n\to\infty$):** By (4), 
\begin{equation*}
	1(Y_n\ne0\}=\sum_{k\in\N}1(S_k<n\le F_k) \tag{6}
\end{equation*}
and hence 
\begin{equation*}
	P(Y_n\ne0\}=\sum_{k\in\N}P(S_k<n\le F_k).  \tag{7}
\end{equation*}
Here and in what follows, $n$ and $k$ are natural numbers. 
Next, consider the duration 
\begin{equation*}
	D_k:=F_k-S_k
\end{equation*}
of the $k$th walk. Since the $R_i$'s are independent and independent of the $T_j$'s, 
the r.v.'s $D_1,D_2,\dots$ are independent and independent of the $T_j$'s, and the $T_j$'s are also independent. So, the r.v.'s $T_1,D_1,T_2,D_2,\dots$ are independent. 
Also, 
\begin{equation*}
	S_k=T_1+\sum_{j=2}^k(D_{j-1}+T_j). \tag{8}
\end{equation*}
So, $S_k$ and $D_k$ are independent and hence   
\begin{equation*}
	P(S_k<n\le F_k)=\sum_{d\in\N}P(D_k=d)P(S_k<n\le S_k+d). 
\end{equation*}
Using (8) and the independence of $T_1,D_1,T_2,D_2,\dots$, for all $d\in\N$ we have 
\begin{equation*}
	P(S_k<n\le S_k+d)\le\sup_{m\in\Z}P(T_k<n\le T_k+d)=\sum_{t=1}^d p_kq_k^{t-1}. 
\end{equation*}
So, 
\begin{equation*}
\begin{aligned}
	P(S_k<n\le F_k)&\le\sum_{d\in\N}P(D_k=d)\sum_{t=1}^d p_kq_k^{t-1} \\ 
	&=\sum_{t\in\N}^d p_kq_k^{t-1}\sum_{d=t}^\infty P(D_k=d) \\ 
	&=\sum_{t\in\N}^d p_kq_k^{t-1}P(D_k\ge t) \\ 
	&=\sum_{t\in\N}^d p_kq_k^{t-1}P(D_1\ge t) \\ 
	&\le\sum_{t\in\N} p_kq_k^{t-1}ct^{-1/2}, 
\end{aligned}
\end{equation*}
since, by (say) the reflection principle, $P(D_1\ge t)\le ct^{-1/2}$ for $t\in\N$. Here and in what follows, $c$ denotes various universal positive real constants. Therefore and because $0<p_k<1/2$, 
\begin{equation*}
\begin{aligned}
	P(S_k<n\le F_k)&\le 2c\sum_{t\in\N} p_k(1-p_k)^t t^{-1/2} \\ 
	&\le 2cp_k \sum_{t\in\N} e^{-p_k t}t^{-1/2} \\ 
	&\le 2cp_k \int_0^\infty e^{-p_k t}t^{-1/2}\,dt =cp_k^{1/2}.  
\end{aligned}
\tag{9}
\end{equation*}

Also, for each $k\in\N$, $P(S_k<n\le F_k)\le P(F_k\ge n)\to0$ (as $n\to\infty$), since $F_k<\infty$ a.s. So, by (7), for each $K\in\N$,
\begin{equation*}
	\limsup_n P(Y_n\ne0\}\le \limsup_n \sum_{k=K}^\infty cp_k^{1/2}.
\end{equation*}
Letting now $K\to\infty$ and recalling (1), we see that $\limsup_n P(Y_n\ne0\}=0$ and hence $Y_n\to0$ in probability (as $n\to\infty$). 

**Noting that $Y_n\not\to0$ a.s. (as $n\to\infty$):** It follows (say by (6)) that $|Y_{S_k+1}|=1$ for all $k\in\N$. Also, $S_k\to\infty$ a.s. as $k\to\infty$. So, $Y_n\not\to0$ a.s. (as $n\to\infty$).