[proofplan] We prove both implications directly from the neighbourhood definition of convergence for nets. The forward implication uses continuity of $f$: every neighbourhood of $f(x)$ pulls back to a neighbourhood of $x$, so eventual membership of $x_i$ in that pullback gives eventual membership of $f(x_i)$ in the original neighbourhood. The reverse implication applies the same argument to the inverse homeomorphism $f^{-1}:Y \to X$. [/proofplan] [step:Use continuity of $f$ to preserve the limit forward] Assume that $x_i \to x$ in $X$. Let $V \subset Y$ be a neighbourhood of $f(x)$ in $Y$. Here $f^{-1}(V)$ denotes the preimage of $V$ under the map $f$, not the inverse map applied to $V$. Since $f$ is continuous, $f^{-1}(V)$ is a neighbourhood of $x$ in $X$. By convergence of the net $(x_i)_{i \in I}$ to $x$, there exists $i_0 \in I$ such that for every $i \in I$ with $i_0 \preceq i$, one has $x_i \in f^{-1}(V)$. Therefore, for every such $i$, \begin{align*} f(x_i) \in V. \end{align*} Since $V$ was an arbitrary neighbourhood of $f(x)$, the net $(f(x_i))_{i \in I}$ converges to $f(x)$ in $Y$. [guided] Assume that $x_i \to x$ in $X$. To prove that $f(x_i) \to f(x)$ in $Y$, we must check the defining neighbourhood condition for the image net: every neighbourhood of $f(x)$ must eventually contain $f(x_i)$. Let $V \subset Y$ be a neighbourhood of $f(x)$ in $Y$. The natural way to use the known convergence in $X$ is to convert this neighbourhood of $f(x)$ into a neighbourhood of $x$. Since $f$ is continuous, the preimage of $V$ under $f$ is a neighbourhood of $x$ in $X$. We write this preimage as $f^{-1}(V)$; in this sentence, $f^{-1}(V)$ means $\{z \in X : f(z) \in V\}$, not evaluation of the inverse homeomorphism. Because $x_i \to x$ in $X$ and $f^{-1}(V)$ is a neighbourhood of $x$, the definition of net convergence gives an index $i_0 \in I$ such that whenever $i_0 \preceq i$, we have \begin{align*} x_i \in f^{-1}(V). \end{align*} By the definition of preimage, this is equivalent to \begin{align*} f(x_i) \in V. \end{align*} Thus every neighbourhood $V$ of $f(x)$ eventually contains the net values $f(x_i)$. This is exactly the statement that $f(x_i) \to f(x)$ in $Y$. [/guided] [/step] [step:Use continuity of the inverse homeomorphism to preserve the limit backward] Assume that $f(x_i) \to f(x)$ in $Y$. Define \begin{align*} g: Y \to X \end{align*} to be the inverse homeomorphism $g=f^{-1}$. Define the identity map \begin{align*} \operatorname{id}_X: X \to X \end{align*} by $\operatorname{id}_X(z)=z$ for every $z \in X$. Let $U \subset X$ be a neighbourhood of $x$ in $X$. Since $g$ is continuous and $g(f(x))=x$, the preimage $g^{-1}(U)$ is a neighbourhood of $f(x)$ in $Y$. By convergence of the net $(f(x_i))_{i \in I}$ to $f(x)$, there exists $i_0 \in I$ such that for every $i \in I$ with $i_0 \preceq i$, one has $f(x_i) \in g^{-1}(U)$. Hence $g(f(x_i)) \in U$. Since $g \circ f=\operatorname{id}_X$, this gives $x_i \in U$ for every $i \in I$ with $i_0 \preceq i$. Since $U$ was an arbitrary neighbourhood of $x$, the net $(x_i)_{i \in I}$ converges to $x$ in $X$. [/step] [step:Conclude the equivalence] The first step proves \begin{align*} x_i \to x \text{ in } X \implies f(x_i) \to f(x) \text{ in } Y. \end{align*} The second step proves the converse implication. Therefore, \begin{align*} x_i \to x \text{ in } X \iff f(x_i) \to f(x) \text{ in } Y. \end{align*} This proves the theorem. [/step]