If $x_n \to 0 \pmod{a}$ for every $a>0$, does it follow that $x_n \to 0$? Let $x_n$ be a real-valued sequence such that $x_n \to 0\pmod{a}$ for every real $a>0$: does it follow that $x_n \to 0$?
Just so there is no ambiguity in the meaning of the question the hypothesis is that
$$ \forall a>0 \; \forall \varepsilon>0 \; \exists n_0\in\mathbb{N} \; \forall n\geq n_0 \; (x_n \in \mathopen]-\varepsilon,+\varepsilon\mathclose[+a\mathbb{Z})$$
where
$$\mathopen]-\varepsilon,+\varepsilon\mathclose[+a\mathbb{Z} := \{x\in\mathbb{R} \; : \; \exists u\in\mathbb{R}\; \exists k\in\mathbb{Z} \; (|u|<\varepsilon \; \land \; x=u+ka)\}$$
(Edit: The question was initially posed with the notation “$A$”: I changed this to $a$ to avoid a notation clash with the accepted answer which uses $A$ for something else (essentially the set of those $a$).)

Addendum: The question has now been answered satisfactorily, but for completeness of MSE, let me still add some remarks that might help explain both whence the question is coming from and what the answer tells us.
Suppose we're given a single $a>0$, say $a>1$ for definiteness, and we want to find a sequence $x_n$ such that $x_n \to 0 \pmod{1}$ and $x_n \to 0 \pmod{a}$ yet $x_n \to +\infty$ in $\mathbb{R}$; in fact, let's even demand $x_n \equiv 0 \pmod{1}$ (i.e., $x_n\in\mathbb{Z}$).  Here's how one can do it: if $a$ is rational it's obvious (just take $x_n$ to be ever larger integers that are also multiples of $a$); and if $a$ is irrational, let $p_n/q_n$ be the convergents of its continued fraction: they satisfy $|q_n a-p_n| < \frac{1}{q_n}$, which means that $p_n$ is both an integer, and also ever closer to $0$ mod $a$, so we have $p_n \equiv 0 \pmod{a}$ and $p_n \to 0 \pmod{a}$, yet $p_n \to +\infty$ in $\mathbb{R}$, as promised. ∎
(For example, the sequence 1, 3, 7, 17, 41, 99, etc. tends to $0$ mod $\sqrt{2}$, and is, of course, identically $0$ mod $1$.)
I suspect (thought I didn't check the details) that standard results on simultaneous Diophantine approximation (maybe applied to the $1/a$: my $a$ is in some sense “backwards” in the question) will similarly give us a sequence $x_n$ tending to $+\infty$ in $\mathbb{R}$ but tending to $0$ mod finitely many $a$'s (and exactly $0$ mod one).  The question was whether this “finitely many $a$'s” can be strengthened to “all $a>0$”, and the accepted answer shows that it's not even possible for a dense $G_\delta$ of $a$.
 A: Foys's answer is good. Here is a different proof using the Fourier transform.

Proof. Observe that the following two facts are equivalent:

*

*$x_n\to 0 \,(\textrm{mod} \,a)$ for every $a\in \mathbb{R}$;

*$e^{-2\pi ix_n \,t}\to 1$ for every $t\in \mathbb{R}.$
Now for every $f\in L^1(\mathbb{R})$, it follows from Lebesgue's dominated convergence theorem that $$\widehat{f}(x_n)=\int_{-\infty}^{+\infty} f(t)e^{-2\pi i x_n \,t}dt \to\int_{-\infty}^{+\infty}f(t)dt=\widehat{f}(0).$$
Take $f(t)=e^{-t^2}$, then $\widehat{f}(x)=\sqrt{\pi}e^{-\pi^2\,x^2}$. The conclusion follows since $\widehat{f}$ is peak at $x=0$.
A: $\newcommand{\N}{\mathbb N}$ The result is obvious if $n\mapsto x_n$ is bounded. We assume it is not. It is then possible to find a subsequence of $n\mapsto x_n$ which converges towards either $+\infty$ or $-\infty$. So without losing generality and by symmetry we can assume that $x_n \underset{n\to +\infty}{\longrightarrow} +\infty$.
It is possible (by picking a further subsequence) to assume that $x_{n+1} - x_n > 1$ for every $n\in \N$. Let $V:= \bigcup_{n\in \N} ]\frac {x_n+x_{n+1}}{2} - \frac {1}{4}; \frac{x_{n+1} + x_n}{2} + \frac 1 4[$ (EDIT:the previous set doesn't seem to work, use $V:=\bigcup_{n\in \N} ]x_n + \frac 1 4; x_n + \frac 3 4[$ instead).
We can then use the following lemma (which immediately gives a contradiction when applied to $V$ with $x>1$): for every unbounded open set $W$ in $]0,+\infty[$, there exists a dense subset $A\subseteq ]0,+\infty[$ such that for every $x\in A$, $\N x \cap W$ is infinite (i.e. unbounded).
The lemma is proven with the Baire theorem. Let $n\in \N$, and $A_n:= \{x \in ]0,+\infty[ \mid \exists k\geq n, kx \in W\}$. Then $A_n$ is an open subset in $]0,+\infty[$. Let us show it is dense in $]0,+\infty[$. Let $x,\varepsilon \in ]0,+\infty[$; there is $M>0$ such that $[M,+\infty[ \subseteq \bigcup_{n\in \N \backslash \{0\}} ]n(x-\varepsilon); n(x+\varepsilon)[$ (these intervals eventually overlap: draw a picture). Since $W$ is unbounded, there exists $u \in (W \backslash \bigcup_{k=0}^n ]k(x- \varepsilon), k(x+\varepsilon)[ ) \cap [M,+\infty[$. Then there exists $p\in\N$ such that $u \in ]p(x-\varepsilon), p(x+\varepsilon)[$ and we have $p>n$, which proves that $]x-\varepsilon, x+\varepsilon[ \cap A_n \neq \emptyset$. The intersection $\bigcap_{n\in \N} A_n$ is dense (Baire) and has the required property.
