Yes, this is the standard way. With nets we often see this "product trick" to construct subnets. Also, using reversely ordered neighbourhoods is a common theme as well. You could compare your write-up with mine here, e.g., and see that is essentially the same.
a. A cluster point of the net $(x_a)_{a \in A}$ in $X$ is a $p$ such that for every (open) neighbourhood $O$ of $p$ and every $a \in A$ there is some $a' \ge a$ such that $x_{a'} \in O$. (The net is frequently in every neighbourhood of $p$). This is probably what you denote by $(x_a)_{a \in A} \dashv p$.
b. It is well-known (e.g. Willard, chapter 11) that $p$ is a cluster point of a net iff there is a subnet of that net that converges to $p$. You seem to assume this fact as known.
c. To a net we associate its tail filter (as Willard also does in chapter 12) and $p$ is a cluster point (or adherence point) of the tail filter iff $p$ is a cluster point of the original net. This is an easy exercise in definitions.
d. Similarly we can define a net $N_{\mathcal{F}}$ from a filter $\mathcal{F}$ as you do (Willard chapter 12 construction again) and note that $p$ is a cluster point of that $N_{\mathcal{F}}$ iff $p$ is a cluster point of $\mathcal{F}$, again an easy exercise in definitions.
So assuming you know
- $X$ is compact iff every filter on $X$ has a cluster point.
We can show the required
- $X$ is compact iff every net has a convergent subnet.
using these correspondences and facts:
$2$, $\Rightarrow$: let $(x_a)_{a \in A}$ be a net in $X$ and $X$ compact. Its tail filter has a cluster point by "$1$, $\Rightarrow$" and that cluster point is also one for the net by c. Then b. tells us that $(x_a)_{a \in A}$ has a convergent subnet.
$2$, $\Leftarrow$: let $\mathcal{F}$ be a filter on $X$ (On $X$ we assume that every net has a convergent subnet), then $N_{\mathcal{F}}$ has a convergent subnet to some $p$. So by b. (reverse direction) $p$ is a cluster point of $N_{\mathcal{F}}$ and so by d. $p$ is a cluster point of $\mathcal{F}$. Then $1$,$\Leftarrow$ tells us that $X$ is compact (as the filter was arbitrary).
So your argument is in essence correct. I just made all the known facts more explicit. So if a-d are all known to you you can use the final proof; maybe you need more details filled in for d? You seem to skip over some details there.
Best Answer
Let $U$ be a neighbourhood of $x$. As $x_\alpha\to x$, there exists an index $\beta\in J$ such that $\alpha\geq\beta \rightarrow x_\alpha\in U$. As the image $g(K)$ is cofinal in $J$, there exists $\theta \in K$ such that $g(\theta)=\beta'\geq \beta$. Then $\zeta\ge\theta\rightarrow g(\zeta)\geq g(\theta)=\beta'\ge\beta$, hence $\zeta \geq\theta \rightarrow f(g(\zeta))=x_{g(\zeta)}\in U$, which means the subnet $f\circ g:K\rightarrow X$ converges to $x$.