No. For example, let $X = \omega_1 + 1$. Then the net $K$ consisting of all countable successor ordinals is convergent, namely to $\omega_1$. But for any subnet $K’ \subset K$, $K’ \cup \{\omega_1\}$ is not compact. Indeed, if it were compact, then it would be closed, so $K’$ would be closed in $\omega_1$. $K’$, being a subnet of the unbounded $K$, must also be unbounded. But any closed, unbounded subset of $\omega_1$ contains some limit ordinals, while $K’$ does not by definition. This is a contradiction.
Let me also note that the statement is not even true for first countable spaces. The problem here is that, while in a first countable space we can certainly find a sequence of elements in the net which converges to the given limit point, there is no reason the said sequence will be a subnet of the original net. Here is an explicit first countable (in fact, even metrizable) counterexample: Let $X = \mathbb{R}^2$ equipped with the post office metric, i.e.,
$$d(x, y) = \begin{cases}
0 &, \text{ if }x = y\\
\|x\| + \|y\| &, \text{ otherwise}
\end{cases}$$
It is not hard to see that compact subsets of $X$ are all countable. Indeed, if $K \subset X$ is compact, then $\{x \in K: \|x\| < \frac{1}{n}\}$ is an open subset of $K$ s.t. every point in $K$ outside of it is isolated. Whence, compactness implies only finitely many elements of $K$ can have norms at least $\frac{1}{n}$. Then, simply take the union over all $n \in \mathbb{N}$ and add the origin to see that $K$ is countable.
Now, let $S^1$ be the unit circle, equipped with a well-ordering $\preceq$ of order type continuum. Let $I = S^1 \times \mathbb{N}$, equipped with the ordering,
$$(x, n) \leq (y, m) \Leftrightarrow x \preceq y \text{ and } n \leq m$$
For each $(x, n) \in I$, let $p_{(x, n)} = \frac{1}{n}x$. Then $p_{(x, n)} \to 0$. But any subnet of $I$ is uncountable, as the cofinality of the continuum is uncountable. Since $I \ni (x, n) \mapsto p_{(x, n)} \in X$ is injective, this implies any subnet must have uncountable image, so no subnet union the limit point $0$ can be compact.
Just for the sake of completeness, let me point out that even $X = [0, 1]$ with its usual topology, which is compact and metrizable (and thus also second countable), does not satisfy the required property. Indeed, pick a subset $A$ of $[\frac{1}{2}, 1]$ of cardinality $\aleph_1$ with the property that any compact subset of $A$ is countable. (This is always possible. Indeed, any uncountable closed subset of the real line has cardinality continuum, so if the continuum hypothesis does not hold, any subset of cardinality $\aleph_1$ would work. In general, any Bernstein set in $[\frac{1}{2}, 1]$ would work.) Equip $A$ with a well-ordering $\preceq$ of order type $\omega_1$. Let $I = A \times \mathbb{N}$, equipped with the ordering,
$$(x, n) \leq (y, m) \Leftrightarrow x \preceq y \text{ and } n \leq m$$
For each $(x, n) \in I$, let $p_{(x, n)} = \frac{1}{3^n}x$. Then $p_{(x, n)} \to 0$. Any subnet of $I$ is uncountable, as the cofinality of $\omega_1$ is uncountable. Since $I \ni (x, n) \mapsto p_{(x, n)} \in X$ is injective, this implies any subnet must have uncountable image. Suppose $K$ is the image of a subnet and $K \cup \{0\}$ is compact. Then $K_n = K \cap [\frac{1}{2 \cdot 3^n}, \frac{1}{3^n}]$ is compact. As it is contained in $\frac{1}{3^n} \cdot A$, by the choice of $A$ we have $K_n$ is countable. But $K = \{0\} \cup (\bigcup_n K_n)$, so $K$ is countable, a contradiction.
