Prove that if a sequence $(x_i)$ does not converge to $a$ in a metric space, there exists neighborhood $A$ of $a$ and $(x_i)\nsubseteq A$.
Further down I present my attempt. I appreciate all input/corrections and welcome any intuition about this problem!
Proof. Since the sequence $(x_i)$ does not converge to $a$, we therefore by definition conclude that it does not exists $\varepsilon$ such that $D(x_i,a)<\varepsilon$. In other words, $D(x_i,a)\geq \varepsilon$.
Since the distance between $x_i$ and $a$ is greater or equal to $\varepsilon$, we can construct an open ball $B_{\varepsilon}(a)$ around $a$ with a radius less that $\varepsilon$. Hence,
$B_{\varepsilon}(a)=\left\{x \in M : d_M(x,a)<\varepsilon \right\}$ where $M$ denotes the metric space.
A subset $U$ of a metric space $(M, d_M)$ is called open if, given any point $x$ in $U$, there exists a real number $\varepsilon > 0$ such that, given any point $y$ in $M$ with $d_M(x, y) < \varepsilon$, $y$ also belongs to $U$. Equivalently, $U$ is open if every point in $U$ has a neighborhood contained in $U$.
According to above, we conclude that there exists a neighborhood $A$ of $a$, since the criterias are met for $B_{\varepsilon}(a)$.
Since all the $x_j$ in $(x_i)$ has a distance $D(x_i,a)\geq \varepsilon$ we can conclude that no $x_j$ can lie within $B_{\varepsilon}$. Thus, any subsequence will also not lie within $B_{\varepsilon}$ and further lying outside the neighborhood $A$ of $a$.
(QED?)
Your statement is clearly false as stated: if we consider (in the reals) the sequence $x_n = \frac{1}{n}$ which converges (only) to $0$, we have that for any $x_n$ from the sequence (e.g. $x=1$, or $x=\frac12$) we cannot have a neighbourhood of that point that misses all points of the sequence, because by definition that neighbourhood contains that $x_n$.
Also, if we take an enumeration of the rationals $\mathbb{Q}$ as a sequence (which we can do as that set is countable), all neighbourhoods of every $x \in \mathbb{R}$ contain infinitely many points of the sequence, yet the sequence does not converge to any point of $\mathbb{R}$.