Proof of Restriction of a Continuous Map, and Continuous Images of Compact Subsets

Claim 1. Let $V\in\mathcal{T}_{Y}$. By the definition of $f|_{A}$,

$$\{x\in A: f|_{A}(x)\in V\}=\{x\in A:f(x)\in V\}=A\cap f^{-1}(V),\qquad f^{-1}(V)=\{x\in X:f(x)\in V\}.$$

Since $f$ is continuous, $f^{-1}(V)\in\mathcal{T}_{X}$, and therefore $A\cap f^{-1}(V)\in\mathcal{T}_{A}$ by the definition of the subspace topology. As $V$ was an arbitrary member of $\mathcal{T}_{Y}$, the map $f|_{A}$ is continuous from $(A,\mathcal{T}_{A})$ to $(Y,\mathcal{T}_{Y})$.

Claim 2. By the definition of a compact subset, the hypothesis that $A$ is compact in $X$ says exactly that the topological space $(A,\mathcal{T}_{A})$ is compact. By claim 1 the map $f|_{A}:(A,\mathcal{T}_{A})\to(Y,\mathcal{T}_{Y})$ is continuous, so Continuous Image of a Compact Space is Compact, applied to $f|_{A}$ in place of the map there and to $(A,\mathcal{T}_{A})$ in place of its domain, shows that the image

$$\{f|_{A}(x):x\in A\}=f(A)$$

is compact in $Y$.