It seems that Gutik's hedgehog is a required counterexample.
I recall that Gutik's hedgehog is the set $$X=\{(0,0)\}\cup\{(\tfrac1n,0):n\in\mathbb N\}\cup\{(\tfrac1n,\tfrac1{nm}):n,m\in\mathbb N\}$$
endowed with the topology $\tau$ consisting of subsets $U\subseteq X$ satisfying the following two conditions:
(1) if $(\frac1n,0)\in U$ for some $n\in\mathbb N$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $k\ge m$;
(2) if $(0,0)\in U$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $n\ge m$ and all $k\in\mathbb N$
It is easy to see that Gutik's hedgehog is countable, second-countable, and Hausdorff but not regular.
In Gutik's hedgehog consider the subset $A=\{(0,0)\}\cup\{(\frac 1n,\frac1{nm}):n,m\in\mathbb N\}$.
It is easy to see that every nontrivial sequence in $A$ has a subsequence that converges in $X$ (to a point in the set $\{(0,0)\}\cup\{(\frac1n,0):n\in\mathbb N\}$). On the other hand, no subspace of $X$ containing $A$ is compact. So, the property (2) is satisfied but (1) does not.
The conditions(2) and (3) are equivalent for sequential spaces. This equivalence follows from two lemmas.
Lemma 1. Let $A$ be a subset of a topological space $X$ such that every open cover of $X$ contains a finite subfamily covering $A$. Then every sequence in $A$ has an accumulation point in $X$.
Proof. Assuming that $A$ contains a sequence $(a_n)_{n\in\omega}$ without accumulation points in $X$, for every $x\in X$, we can find a neighborhood $U_x\subseteq X$ of $X$ such that the set $\{n\in\omega:a_n\in U_x\}$ is finite. Then for every finite subfamily $\mathcal F$ of the open cover $\{U_x:x\in X\}$ of $X$ the set $\{n\in\omega:a_n\in \bigcup\mathcal F\}$ is finite and hence $\{a_n:n\in\omega\}\not\subseteq\bigcup\mathcal F$ and $A\not\subseteq \bigcup\mathcal F$. $\quad \square$
Lemma 2. If a sequence $(x_n)_{n\in\omega}$ in a sequential space $X$ has an accumulation point in $X$, then $(x_n)_{n\in\omega}$ has a subsequence that converges to some point of $X$.
Proof. Let $a$ be an accumulation point of the sequence $(x_n)_{n\in\omega}$ in $X$.
For every point $x\in X$ let $I_x$ be the intersection of all neighborhoods of $x$ in $X$. It is easy to see that for every $x\in X$ and $y\in I_x$ we have $I_y\subseteq I_x$.
For a subset $A\subseteq X$, consider the set $\dot A:=\{x\in X:I_x\cap A\ne\emptyset\}$ and observe that $A\subseteq \dot A$.
If for some $x\in X$ the set $\Omega_x:=\{n\in\omega:x_n\in I_x\}$ is infinite, then $(x_n)_{n\in\Omega_x}$ is a required convergent (to $x$) subsequence of the sequence $(x_n)_{n\in\omega}$ and we are done. So, assume that for every $x\in X$ the set $\Omega_x$ is finite. In this case, the set $A=\{x_n:n\in\omega\setminus\Omega_a\}$ is not closed in $X$ as $a\in\overline{A}\setminus A$.
Moreover, the set $\dot A$ is not closed as $A\subseteq \dot A\subseteq\bar{A}$ and $a\notin \dot A$. By the sequentiality of $X$, the set $\dot A$ contains a sequence $(a_k)_{k\in\omega}$ that converges to some point $b\in X\setminus \dot A$. If $a_k\in I_b$ for some $k\in\omega$, then $I_{a_k}\subseteq I_b$ and $a_k\in \dot A$ imply $I_b\cap A\supseteq I_{a_k}\cap A\ne\emptyset$ and hence $b\in \dot A$, which contradicts the choice of $b$. This contradiction implies that $I_b\cap\{a_k:k\in\omega\}=\emptyset$ and hence the set $\{a_k:k\in\omega\}$ is infinite. Since $\{a_k:k\in\omega\}\subseteq \dot A$, for every $k\in\omega$ there exists a number $i_k\in\omega$ such that $x_{i_k}\in I_{a_k}$. We claim that $b$ is the limit of the sequence $\{x_{i_k}:k\in\omega\}$. Indeed, for every open neighborhood $O_b$ of $b$ in $X$, there exists $m\in\omega$ such that $\{a_k:k\ge m\}\subseteq O_b$. Then for every $k\ge m$ we have $x_{i_k}\in I_{a_k}\subseteq O_b$, which means that the sequence $(x_{i_k})_{k\in\omega}$ converges to $b$.
Assuming that the set $F=\{x_{i_k}:k\in\omega\}$ is finite and taking into account that $b\notin \dot A$, we can find a neighborhood $O_b$ such that $O_b\cap F=\emptyset$, which contradicts the convergence of the seqeunce $(x_{i_k})_{k\in\omega}$ to $b$. This contradiction shows that the set $F$ is infinite. Then we can choose an increasing number sequence $(n_k)_{k\in\omega}$ such that $F=\{x_{n_k}\}_{k\in\omega}$ and $x_{n_i}\ne x_{n_j}$ for $i\ne j$. For every $k\in\omega$ find $p_k\in\omega$ such that $n_k=i_{p_k}$. The convergence of the sequence $(x_{i_p})_{p\in\omega}$ to $b$ implies the convergence of the sequence $(x_{n_k})_{k\in\omega}=(x_{i_{p_k}})_{k\in\omega}$ to $b$. Therefore, $(x_{n_k})_{k\in\omega}$ is a required convergent subsequence of the sequence $(x_n)_{n\in\omega}$. $\quad\square$
