Inclusion of a subspace is continuous

Statement

Let $X$ be a topological space, let $S \subseteq X$ have the subspace topology and let $\iota \colon S \to X$ be the inclusion map $\iota(s) = s$. Then $\iota$ is continuous.

Remarks

Proof: for $U$ open in $X$, the preimage is $\iota^{-1}(U) = S \cap U$, which is open in $S$. Thus the restriction $f|_S = f \circ \iota$ of a continuous map $f \colon X \to Y$ is continuous. A map $g \colon Z \to S$ is continuous if and only if $\iota \circ g \colon Z \to X$ is continuous. This is the universal property of the subspace topology.

Explanations

L1 human

Putting a subspace back into the big space is a continuous map. So when you restrict a continuous function to a subset, you get a continuous function.