[proofplan] We prove both implications directly. Continuity sends neighbourhoods of $f(x)$ back to neighbourhoods of $x$, so any sequence converging to $x$ is eventually mapped into each neighbourhood of $f(x)$. Conversely, if $f$ were not continuous, then some [open set](/page/Open%20Set) $V\subset Y$ would have a non-open preimage; first-countability lets us choose a sequence converging to a point of that preimage while staying outside the preimage, contradicting the assumed sequential property. [/proofplan]

[step:Use continuity to transfer eventual membership through preimages] Assume that $f:X\to Y$ is continuous. Let $x\in X$, and let $(x_k)_{k\in\mathbb N}$ be a sequence in $X$ such that $x_k\to x$ in $(X,\tau_X)$. To prove $f(x_k)\to f(x)$ in $(Y,\tau_Y)$, let $V\in\tau_Y$ be an open neighbourhood of $f(x)$. Since $f$ is continuous, the preimage \begin{align*} f^{-1}(V):=\{z\in X:f(z)\in V\} \end{align*} belongs to $\tau_X$. Also $x\in f^{-1}(V)$ because $f(x)\in V$. Thus $f^{-1}(V)$ is an open neighbourhood of $x$ in $X$. Since $x_k\to x$, there exists $N\in\mathbb N$ such that for every $k\in\mathbb N$ with $k\ge N$, one has $x_k\in f^{-1}(V)$. Equivalently, for every $k\ge N$, \begin{align*} f(x_k)\in V. \end{align*} Because $V$ was an arbitrary open neighbourhood of $f(x)$, this proves $f(x_k)\to f(x)$ in $Y$. [/step]

[step:Construct a convergent sequence witnessing failure of continuity]Assume conversely that for every $x\in X$ and every sequence $(x_k)_{k\in\mathbb N}$ in $X$, convergence $x_k\to x$ implies $f(x_k)\to f(x)$. We prove that $f$ is continuous. Let $V\in\tau_Y$ be open. We show that $f^{-1}(V)$ is open in $X$. If $f^{-1}(V)=\varnothing$, then $f^{-1}(V)$ is open, so suppose $f^{-1}(V)\ne\varnothing$ and let $x\in f^{-1}(V)$. Since $X$ is first-countable, there exists a countable local base $(U_m)_{m\in\mathbb N}$ at $x$, meaning that each $U_m$ is an open neighbourhood of $x$ and every open neighbourhood of $x$ contains some $U_m$. Define a new sequence of neighbourhoods $(B_m)_{m\in\mathbb N}$ by \begin{align*} B_m:=\bigcap_{j=1}^{m} U_j. \end{align*} Each $B_m$ is an open neighbourhood of $x$, since it is a finite intersection of open neighbourhoods of $x$. Suppose, for contradiction, that $f^{-1}(V)$ is not a neighbourhood of $x$. Then no open neighbourhood of $x$ is contained in $f^{-1}(V)$. In particular, for every $m\in\mathbb N$, \begin{align*} B_m\not\subset f^{-1}(V). \end{align*} Hence for each $m\in\mathbb N$ we may choose a point $x_m\in B_m\setminus f^{-1}(V)$. This defines a sequence $(x_m)_{m\in\mathbb N}$ in $X$. We claim that $x_m\to x$ in $X$. Let $W\in\tau_X$ be an open neighbourhood of $x$. Since $(U_m)_{m\in\mathbb N}$ is a local base at $x$, there exists $j\in\mathbb N$ such that $U_j\subset W$. For every $m\ge j$, the definition of $B_m$ gives \begin{align*} x_m\in B_m\subset U_j\subset W. \end{align*} Thus $(x_m)$ is eventually in every open neighbourhood $W$ of $x$, so $x_m\to x$. By the assumed sequential property, $f(x_m)\to f(x)$ in $Y$. Since $x\in f^{-1}(V)$, we have $f(x)\in V$, so $V$ is an open neighbourhood of $f(x)$. Therefore convergence $f(x_m)\to f(x)$ implies that there exists $N\in\mathbb N$ such that for every $m\ge N$, \begin{align*} f(x_m)\in V. \end{align*} Equivalently, $x_m\in f^{-1}(V)$ for every $m\ge N$. This contradicts the construction $x_m\notin f^{-1}(V)$ for every $m\in\mathbb N$. Therefore $f^{-1}(V)$ is a neighbourhood of each of its points. Hence $f^{-1}(V)$ is open in $X$. Since $V\in\tau_Y$ was arbitrary, the preimage of every open subset of $Y$ is open in $X$, and therefore $f$ is continuous.[/step]

[guided]We assume the sequential condition and prove continuity by the open-preimage criterion: for every open set $V\in\tau_Y$, we must prove that $f^{-1}(V)$ is open in $(X,\tau_X)$. Fix such a set $V$. If $f^{-1}(V)=\varnothing$, then $f^{-1}(V)$ is open by the topology axioms, so we may assume $f^{-1}(V)\ne\varnothing$ and fix a point $x\in f^{-1}(V)$. The goal is to prove that $f^{-1}(V)$ is a neighbourhood of this arbitrary point $x$. Since $(X,\tau_X)$ is first-countable, there exists a countable local base $(U_m)_{m\in\mathbb N}$ at $x$. This means that each $U_m\in\tau_X$ contains $x$, and for every open neighbourhood $W\in\tau_X$ of $x$, there exists $j\in\mathbb N$ such that $U_j\subset W$. Define the decreasing neighbourhood sequence $(B_m)_{m\in\mathbb N}$ by \begin{align*} B_m:=\bigcap_{j=1}^{m}U_j. \end{align*} Each $B_m$ is an open neighbourhood of $x$, because it is a finite intersection of open sets containing $x$. Suppose, toward a contradiction, that $f^{-1}(V)$ is not a neighbourhood of $x$. Then no open neighbourhood of $x$ is contained in $f^{-1}(V)$. Applying this to the open neighbourhood $B_m$ for each $m\in\mathbb N$, we get \begin{align*} B_m\not\subset f^{-1}(V). \end{align*} Thus for every $m\in\mathbb N$ there exists a point $x_m\in B_m\setminus f^{-1}(V)$. This choice defines a sequence $(x_m)_{m\in\mathbb N}$ in $X$. We now verify that this sequence converges to $x$. Let $W\in\tau_X$ be an open neighbourhood of $x$. Since $(U_m)_{m\in\mathbb N}$ is a local base at $x$, choose $j\in\mathbb N$ such that $U_j\subset W$. For every $m\ge j$, the definition of $B_m$ gives $B_m\subset U_j$, because $U_j$ is one of the finitely many sets intersected in $B_m$. Hence, for every $m\ge j$, \begin{align*} x_m\in B_m\subset U_j\subset W. \end{align*} Therefore $(x_m)_{m\in\mathbb N}$ is eventually in every open neighbourhood of $x$, which is exactly the definition of $x_m\to x$ in the [topological space](/page/Topological%20Space) $(X,\tau_X)$. By the assumed sequential property, the convergence $x_m\to x$ implies $f(x_m)\to f(x)$ in $(Y,\tau_Y)$. Since $x\in f^{-1}(V)$, we have $f(x)\in V$, so $V$ is an open neighbourhood of $f(x)$. The definition of convergence in $Y$ therefore gives an index $N\in\mathbb N$ such that for every $m\in\mathbb N$ with $m\ge N$, \begin{align*} f(x_m)\in V. \end{align*} Equivalently, $x_m\in f^{-1}(V)$ for every $m\ge N$. This contradicts the construction $x_m\in B_m\setminus f^{-1}(V)$ for every $m\in\mathbb N$. The contradiction shows that $f^{-1}(V)$ is a neighbourhood of $x$. Because $x\in f^{-1}(V)$ was arbitrary, $f^{-1}(V)$ is a neighbourhood of each of its points. A subset of a topological space is open exactly when it is a neighbourhood of each of its points, so $f^{-1}(V)$ is open in $X$. Since $V\in\tau_Y$ was arbitrary, the preimage of every open subset of $Y$ is open in $X$, and therefore $f:X\to Y$ is continuous.[/guided]