IMHO, you have to work with some subnet constructions, to get a feel for them.
A subnet is a function on index sets, as a net is a function from an index set into a space, as you probably know.
So if $f: (I, \le) \rightarrow X$ ,is a net (a sequence is just a net from $(\mathbb{N}, \le)$, where $(I, \le)$ is a directed set ($i \le i$, $i \le j \le k \rightarrow i \le k$, $\forall i_1, i_2 \in I \exists j \in I: i_1 \le j, i_2 \le j$) then a net $g:(J, \le) \rightarrow X$ is a subnet of $f$ iff there exists $h: J \rightarrow I$ that is order preserving ($j_1 \le j_2 \rightarrow h(j_1) \le h(j_2)$ and which has a cofinal image ($\forall i \in I, \exists j \in J: h(j) \ge i$) such that $f \circ h = g$.
There are other notions of subnet that are not equivalent, see this question and answer, but I find this one the easiest to understand, as it looks the most like a traditional subsequence. It's in fact exactly a subsequence if we'd demand $J = \mathbb{N}$ as well. But $J$ can be much bigger.
E.g. in your example you use the map $\mathbb{R}^+ \rightarrow \mathbb{N}$ defined by $x \rightarrow \lfloor x \rfloor$, which satisfies the requirements if both sets have their usual orders.
To get a subnet converging to $a$, we usually take a directed set that is related to $a$: let the directed set be $I =\mathscr{N}_a$ all open neighbourhoods of $a$, ordered by reverse inclusion $i_1 \le i_2$ iff $i_2 \subseteq i_1$. Directedness follows from standard inclusion properties and the fact that the intersection of two neighbourhoods of $a$ is again a neigbourhood of $a$.
Then we use that $a$ is a limit point (every neighbourhood of $a$ contains infinitely many points of the sequence, really an accumulation point) of $\{x_n: n \in \mathbb{N}\}$ to make a subnet: for each $U \in \mathscr{N}_a$ define $f(U) = \min \{n: x_n \in U\}$ and set $g(U) = x_{f(U)}$ where $f$ is well-defined as the minimum of a non-empty subset of the (well-ordered) set $\mathbb{N}$. This defines a net $g: I \rightarrow X$, and $f$ shows it's clearly a subnet of the orginal sequence in the above sense (a smaller neighbourhood can only have a larger minimal index, so order is preserved, and the image is cofinal as every neighbourhood must intersect infinitely many elements of the sequence). It also converges to $a$ as any open neighbourhood $O$ of $a$ is in the index set, so we have $x_O \in O$ and whenever $O' \ge O$ we actually have $O' \subseteq O$ so $x_{O'} \in O' \subseteq O$ as well. So every neighbourhood defines it own "tail" sets (all neighbourhoods inside it) in this net of neighbourhoods $I$; the order on $I$ is not linear, that's where you have to broaden your intuition, sets of the form $\{i \in I: i \ge i_0\}$ can be very thin threads in $I$, not "almost all points", like for sequences. So convergence wrt a net on such an order is a bit strange.
If $a$ would have a countable local base we could just take a countable base as $I$ which is decreasing and $I$ just becomes a copy of $\mathbb{N}$ and we get a subsequence instead of a subnet. To get something really different you normally use large products like $[0,1]^\mathbb{R}$ or spaces like $\beta \omega$, where the points usually have quite complicated neighbourhood structures, and many non-compatible neighbourhoods exist (for 2 neighbourhoods one need not be a subset of the other: in metric spaces neighbourhoods are essentially countable linear structures, consider the balls $B(x,\frac{1}{n})$ for fixed $x$, etc. that is why sequences suffices there.)
This question and answer also illustrate nicely how we can not a have subsequence but still a subnet that converges.
