A sequence in a complete metric space whose closed balls are totally bounded, $Y$, is semiconvergent if and only if it has at most one limit point in $Y$.
Proof:
Suppose $(x_n)$ has two limit points in $Y$, then choose a bounded open set containing both. Choose the subsequence of $(x_n)$ that lies in this open set. Now we have a bounded subsequence that necessarily still has two limit points, hence is not convergent.
For the converse, it suffices to show that a bounded sequence, $(x_n)$, with a unique limit point, $x$, is convergent. Since $(x_n)$ is bounded, it is contained in a closed ball, which is compact by total boundedness of closed balls and completeness of $Y$. Call this closed ball $K$. Then if $(x_n)$ doesn't converge to the unique limit point $x$, there is $\epsilon > 0$ such that $(x_n)$ has infinitely many terms not contained in the open ball $U=B_\epsilon(x)$. Then let $(y_n)$ be the subsequence of $(x_n)$ contained in $K\setminus U$, which is a closed and hence compact subset of $K$. Since compactness implies sequential compactness for metric spaces, $(y_n)$ has a limit point in $K\setminus U$, and thus so does $(x_n)$. Contradiction.
Note:
In particular, this applies to $Y=\Bbb{R}^n$ for any $n$.
Best Answer
Let $x$ be real. For each positive integer $k$, there is an integer $m_k$ such that $x-\frac{1}{k}<\frac{m_k}{k+1}<x+\frac{1}{k}$. Now the rational sequence $(\frac{m_k}{k+1})_k$ converges to $x$, so the irrational sequence $(\frac{m_k+\sqrt 2}{k+1})_k$ converges to $x$.
Alternatively, if $x$ is rational, then the rational sequence $(x,x,x,...)$ and the irrational sequence $(x+\frac{\sqrt 2}{k})_k$ converge to $x$. If $x$ is irrational, then the irrational sequence $(x,x,x,...)$ converges to $x$.