Below are the details for the nice example outlined in D. Thomine's comment:
Let $(Y_n)$ be independent r.v. with values in $\{0, 1\}$ and with $\mathbb{P}(Y_n = 0) = 1-1/n$. Also, independently from the $(Y_n)$, let $(Z_n)$ be i.i.d. r.v. with values in $\{-1, 0, 1\}$ according to the limit distribution. Starting from $X_0 = 0$, we iteratively define $X_{n}$ by the following transition rule:
\begin{array}{|c|c|c|}
\hline
& Y_n = 0 & Y_n = 1 \\ \hline
|X_{n-1}| \leq 1 & X_{n-1} & X_{n-1} \pm 3 \\ \hline
|X_{n-1}| > 1 & Z_n & nX_{n-1} - (n-1)Z_n \\ \hline
\end{array}
(Here $\pm 3$ means with equal chance of adding or subtracting 3, independently from everything else.)
It's easy to verify this is a martingale. Also, observe that $|X_n| > 1$ if and only if $Y_n = 1$.
To verify condition (2), notice that $|X_n| > 4$ only if $Y_{n-1} = Y_n = 1$. It follows from the convergence of $\sum \frac{1}{n^2}$ and the Borel-Cantelli Lemma that this a.s. will not happen infinitely often.
To verify condition (1), notice that for $a = -1, 0, 1$, $X_n = a$ if $Y_{n-1} = 1, Z_n = a$, provided $Y_n \neq 1$. If follows from the divergence of $\sum \frac{1}{n}$, the 2nd Borel-Cantelli Lemma, and the previous paragraph that this a.s. will happen infinitely often.
Finally, to verify condition (3), we just compute the distributions of $X_n$. Notice that $X_n$ are integer-valued by construction, and $P(|X_n| > 1) = P(Y_n = 1) = 1-1/n \rightarrow 0$. It remains to show $P(X_n = a) \rightarrow p_a$ for $a = -1, 0, 1$, where $p_a$ are pre-assigned. Using the transition rule, we have
\begin{align}
P(X_n = a) &= P(Y_n = 0) [P(X_{n-1} = a) + P(|X_{n-1}| > 1) P(Z_n = a)] \\
&= (1-1/n) (P(X_{n-1} = a) + p_a/(n-1)).
\end{align}
Define $p_n = \frac{P(X_n = a)}{1-1/n}$, this becomes
\begin{align}
p_n &= [1-1/(n-1)] p_{n-1} + p_a/(n-1) \\
p_n - p_a &= [1-1/(n-1)] (p_{n-1} - p_a) \\
&= \prod_{k=1}^n [1-1/(k-1)] (p_0 - p_a) \\
&\rightarrow 0.
\end{align}
Thus, $P(X_n = a) = (1-1/n) p_n \rightarrow p_a$, and we are done showing condition (3).
