Since I've been chewing on this for a while now, allow me to first construct a divergent sequence $(X_n)_{n\in\mathbb N}$ such that $(e^{itX_n})_{n\in\mathbb N}$ is convergent for all $t\in\mathbb Q$, before I shall solve the original problem below.
Counterexample for $\mathbb Q$ (incomplete space)
Let $(q_n)_{n\in\mathbb N}$ be an enumeration of $\mathbb Q\setminus 0$. Observe that $\bigcap_{k=1}^n q_k\mathbb Z=r_n\mathbb Z$ for some $r_n\in \mathbb Q_{>0}$. By letting $y_1=0$, and selecting $y_n\in r_n\mathbb Q$ with $y_n>y_{n-1}+1$ we obtain a diverging sequence $(y_n)_{n\in\mathbb N}$. If $t\in\mathbb Q$, then $ty_n\in\mathbb Z$ for almost all $n$: If $t=0$, this is trivial; if $t\ne 0$, we find $N$ with $\frac1t=q_N$ and have $y_n\in q_N\mathbb Z$ for all $n\ge N$.
Therefore, by letting $X_n=2\pi y_n$ we find a divergent sequence $(X_n)_{n\in\mathbb N}$ such that $(e^{itX_n})_{n\in\mathbb N}$ is convergent (in fact, is eventually constant $=1$) for all $t\in \mathbb Q$.
Proof for the case $\mathbb R$ (complete space)
What's different with $\mathbb R$ instead of $\mathbb Q$? First let's see that $(X_n)_{n\in\mathbb N}$ is bounded: For $c>0$ and $N\in\mathbb N$ consider the set $A_{c,N}=\{t\in\mathbb R\mid\forall n,m\ge N\colon t(X_n-X_m)\in[-c,c]+2\pi\mathbb Z\}.$ If $t\notin A_{c,N}$, then there are $n,m\ge N$, $k\in\mathbb Z$ with $2k\pi+c, which also holds for $t'$ if
$|t'-t|<\frac{\min\{t(X_n-X_m)-2k\pi-c,2(k+1)\pi-c-t(X_n-X_m)\}}{|X_n-X_m|}.$ Therefore the complements of the $A_{c,N}$ are open and the $A_{c,N}$ themselves are closed. The convergence of $(e^{itX_n})_{n\in\mathbb N}$ for all $t\in\mathbb R$ implies that for any $c>0$ we have $\mathbb R=\bigcup_{N\in\mathbb N}A_{c,N}$ Specifically, we can consider $c=\frac{2\pi}5$. By the Baire category theorem, there exists an $N$ such that $A_{c,N}$ contains an open interval $(t_0-\epsilon,t_0+\epsilon)$ with $\epsilon>0$. Without loss of generality, $t_0\ne0$ and $\epsilon< |t_0|$. Let $M=\max\left\{|X_1|,\ldots,|X_N|\right\}+\frac\pi{\epsilon}.$ Then $(X_n)_{n\in\mathbb N}$ is bounded by $M$. To prove this, assume that $|X_n|>M$ for some $n$. Then clearly $n>N$ and $|X_n-X_N|>\frac\pi{\epsilon}$. Because $n>N$, there exists $k\in\mathbb Z$ such that $|t_0(X_n-X_N)-2k\pi|\le c$. Let $t_1=t_0+\frac\pi{(X_n-X_N)}$ so that $|t_1-t_0|<\epsilon$ and hence $t_1\in A_{c,N}$. Then there is $k'\in\mathbb Z$ with $|t_1(X_n-X_N)-2k'\pi|\le c$. But $|t_1(X_n-X_N)-t_0(X_n-X_N)|=\pi$ makes this impossible because $2c<\pi<2\pi-2c$: the left inequality rules out $k=k'$, the other rules out $|k-k'|\ge 1$. We conclude that $(X_n)_{n\in\mathbb N}$ is bounded.
Now the final step is easy: If $|X_n|\le M$ for all $n$, then consider the case $0. For such $t$, all $tX_n$ are in $(-\frac\pi2,\frac\pi2)$. Since the map $(-\frac\pi2,\frac\pi2)\to \mathbb C$, $x\mapsto e^{ix}$ is an embedding, convergence of $(e^{itX_n})_{n\in\mathbb N}$ implies convergence of $(tX_n)_{n\in\mathbb N}$ and finally convergence of $(X_n)_{n\in\mathbb N}$.$_\blacksquare$