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Every nonempty open set in the given topology contains $1$. Suppose a continuous map $f$ took distinct values $f(m)$ and $f(n)$. Choose disjoint open intervals $U,V\subset \mathbb{R}$ containing these respective values. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty open sets in $\mathbb{N}$, and therefore both contain $1$. This is impossible because the inverse images of disjoint sets are disjoint. Thus every such continuous map is constant. Conversely, constant maps are continuous. This is the continuous real maps from the initial-segment topology argument, using the Hausdorff separation of distinct real points.