Suppose that $a_n=n\bmod3$ for each $n\in\omega$. For $k=0,1,2$ let $A_k=\{n\in\omega:n\bmod 3=k\}$; then $\{A_0,A_1,A_2\}$ is a partition of $\omega$, so exactly one of the sets $A_k$ belongs to $\mathscr{U}$. If $A_k\in\mathscr{U}$, then $\langle a_n:n\in\omega\rangle_{\mathscr{U}}=\mathbf{k}_\mathscr{U}$, where $\mathbf{k}=\langle k,k,k,\ldots\rangle$.
Now suppose that $a_n=\sin n$ for each $n\in\omega$. The sequence $a=\langle a_n:n\in\omega\rangle$ is bounded, so this answer shows that there is a real number $x\in[-1,1]$ such that $x=\mathscr{U}$-$\lim a$, i.e., such that $\{n\in\omega:|a_n-x|<\epsilon\}\in\mathscr{U}$ for each $n\in\omega$. It follows that $a_{\mathscr{U}}$ is infinitesimally close to $\mathbf{x}_{\mathscr{U}}$. Note that this argument works for every bounded sequence of reals.
Since $\mathscr{U}$ is a free (non-principal) ultrafilter, $\bigcap\mathscr{U}=\varnothing$, so your construction is impossible.
The answer to your final question is that it’s not true that every hyperreal is infinitely close to some real: the infinite hyperreals are not. The simplest example of such a hyperreal is $\alpha=\langle n:n\in\omega\rangle_{\mathscr{U}}$. If $\mathbf{x}_{\mathscr{U}}$ is any real in ${^*\Bbb R}$, let $m$ be any integer larger than $|x|+1$, say; then $|n-x|\ge 1$ for $n\ge m$, so $\alpha\ge\mathbf{x}_{\mathscr{U}}+1$, which certainly implies that $\alpha$ is not infinitely close to $\mathbf{x}_{\mathscr{U}}$.
Added: More generally, given any sequence $a=\langle a_n:n\in\omega\rangle$ of reals, if there is a $U\in\mathscr{U}$ such that $\langle a_n:n\in U\rangle$ is bounded, then we can apply the argument of $(2)$ above to see that $a_{\mathscr{U}}$ is infinitely close to some standard real: any two sequences that agree on a member of $\mathscr{U}$ give rise to the same element of ${^*\Bbb R}$.
A hyperreal $\alpha=\langle a_n:n\in\omega\rangle_{\mathscr{U}}$ is infinite if $\mathbf{x}<|\alpha|$ for each $x\in\Bbb R$. Thus, if $\alpha$ is not infinite, there is an $x\in\Bbb R$ such that $|\alpha|\le\mathbf{x}$. Let $U=\{n\in\omega:|a_n|\le x\}$; then $U\in\mathscr{U}$ so $\langle a_n:n\in\omega\rangle$ is bounded on $U$, and $\alpha$ is therefore infinitely close to some standard real.
Conversely, suppose that $x\in\Bbb R$, and $a_{\mathscr{U}}$ is infinitely close to $\mathbf{x}$. Let $U=\{n\in\omega:|a_n-x|<1\}$; then $U\in\mathscr{U}$, and $a$ is bounded on $U$. Thus, $a_{\mathscr{U}}$ is infinitely close to some standard real if and only if the sequence $a$ is bounded on some element of $\mathscr{U}$, and infinite if and only if $a$ is unbounded on every element of $\mathscr{U}$.
In general there is no way to determine the standard part of the hyperreal defined by a given sequence of reals, because there’s no way to pin down exactly which subsets of $\omega$ belong to $\mathscr{U}$. About all that you know for sure (in general) is that $\mathscr{U}$ contains all of the cofinite sets.
