Let $X$ be a Hausdorff space. Assume that $x'\in X$ and $(x_n)$ is a sequence convergent to $x'$ and such that $x_n\neq x'$ for any $n$. Then $\{x_n\}$ as a subspace of $X$ has to be discrete because otherwise it would have a subsequence convergent somewhere else than $x'$ which in Hausdorff spaces is not possible.
I wonder whether we can generalize this do nets. The obvious approach by simply replacing "sequence" with "net" fails, because we can take our sequence $x:\mathbb{N}\to X$ and extend it to a net $x:\mathbb{Z}\to X$ (with the increasing order), where we can put pretty much any value on the negative integers, in particular make the image dense in $X$ if possible. But this our new $x$ net has a subnet with the desired property.
So the question is: let $X$ be a Hausdorff space and $x\in X$ such that there is a net $f:I\to X$ convergent to $x$ with $x\not\in im(f)$. Does there exist a net $g$ convergent to $x$ with $x\not\in im(g)$ such that the image of $g$ is discrete? I do not require $g$ to be a subnet of $f$ but it would be a nice bonus.
Here is a counterexample. Let $X=\mathbb{R}$ with the topology generated by the usual topology together with the cocountable topology. That is, a set $U$ is open iff it has the form $V\setminus A$ where $V$ is open in the usual topology and $A$ is countable.
I claim first that every discrete subset $A\subseteq X$ is countable. Indeed, for each $x\in A$, we can pick a neighborhood $U_x$ of $x$ that isolates $x$ in $A$, which comes from a neighborhood $V_x$ of $x$ in the usual topology which contains only countably many points of $A$. Shrinking each $V_x$, we may assume they are all intervals with rational endpoints, so there are only countably many distinct $V_x$. Thus we have covered $A$ by countably many sets, each of which has countable intersection with $A$, so $A$ is countable.
Now since $X$ has no isolated points, every point $x\in X$ is the limit of some net whose image does not contain $x$. However, suppose $g$ is a net converging to $x$ whose image is discrete. The image of $g$ is then countable, and thus closed, so since $x$ is the limit of $g$, $x$ must be in the image.
(Let me remark that this construction also gives an example in a very nice space (namely, $\mathbb{R}$ with its usual topology) where no $g$ with discrete image exists which is a subnet of $f$. Indeed, let $f$ be a net converging to some point $x$ in the space $X$ above whose image does not contain $x$. Then $f$ also converges to $x$ in the usual topology of $\mathbb{R}$. But $f$ is eventually outside every countable set, so the same is true of any subnet of $f$. Such a subnet thus cannot have image which is discrete.)