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
$$|t'-t|<\frac{\min\{t(X_n-X_m)-2k\pi-c,2(k+1)\pi-c-t(X_n-X_m)\}}{|X_n-X_m|}.$$
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