Is the map continuous?

Consider two topological spaces:

1. The set of positive integers $$\mathbb{N}$$ and the topology $$\mathscr{M}$$ given by $$\emptyset$$ and the subsets of $$\mathbb{N}$$ that contain $$1\in\mathbb{N}$$.
2. The set $$[0,1]\in\mathbb{R}$$ with the metric $$d(t,s)=|t-s|$$.

I want to determine íf the function $$f:[0,1]\rightarrow \mathbb N$$ defined by $$f(t):= \begin{cases}n, & t=0 \\ 1, & t \in] 0,1[ \\ m, & t=1\end{cases}$$ is continuous. I know that $$f$$ is continuous if $$\forall t\in[0,1] \text{ and }\forall U\in\mathfrak{U}(f(t)) :f^{-1}(U)\in\mathfrak{U}(t).$$ So if I can show that $$f^{-1}(U)\in\mathfrak{U}(t)$$ for any $$U\in\mathfrak{U}(n)\cup\mathfrak{U}(1)\cup\mathfrak{U}(m)$$ then I can conclude that $$f$$ is continuous. For $$t=0$$, the neighborhoods $$U\in\mathfrak{U}(f(t))$$ are sets that contain $$n$$ and $$1$$, so $$[0,1)\subseteq f^{-1}(U)$$. This is a neighborhood of $$t=0$$ if there exists an open ball in $$[0,1)$$ that contains $$0$$. However, does this open ball exist when the set is closed from the bottom? And am I approaching the problem of determining if $$f$$ is continuous correctly?

Yes, you are approaching it correctly. Note that, in $$[0,1]$$, $$[0,1)$$ is an open set, since it is the ball centered at $$0$$ with radius $$1$$.
By a similar argument, $$(0,1]$$ is an open subset of $$[0,1]$$, and therefore $$f$$ is continuous at $$1$$ too.