Let \(X\) be a topological space and \(S\subseteq X\) a subspace of \(X\). The map \(\iota_S:S\hookrightarrow X\) denotes the inclusion map of \(S\) into \(X\). For any topological space \(Y\) a function \(f:Y\to S\) is continuous if and only if the composite map \(\iota_S\circ f\colon Y\to X\) is continuous. The subspace topology is the unique topology on \(S\) with this property.
The first statement is an immediate consequence of the identity \(\iota_S^{-1}(V)=V\cap S\).
For uniqueness we consider \(S\) endowed with a topology \(\mathcal{T}_c\) that satisfies the characteristic property and denote the subspace topology of \(S\) with \(\mathcal{T}_s\). Let \(\Id_{sc}\colon (S,\mathcal{T}_s)\to (S,\mathcal{T}_c)\) be the identity map of \(S\). Then \(\iota_c\circ \Id_{sc} =\iota_s\) is continuous, since the inclusion map is continuous according to (0x678ea93e) . With a similar argument we show that \(\Id_{cs}\colon (S,\mathcal{T}_c)\to (S,\mathcal{T}_s)\) is also continuous and therefore the topologies \(\mathcal{T}_s\) and \(\mathcal{T}_c\) are identical.