It is given that the function $f(x) = \sqrt{x}$ is continuous on $[0, \infty)$. I am trying to prove that the function $f(x) = \sqrt{x}$ is also continuous on $[0, 1]$. Can someone please verify my proof? Thanks!
Proof: Let $c \in [0, \infty)$ be arbitrary. We have already established that $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\mid x-c \mid \implies \mid f(x) - f(c) \mid < \epsilon$ where $f: [0, \infty) \rightarrow \mathbb{R}$ and $f(x) = \sqrt{x}$. In other words, we have already established that the function $f(x) = \sqrt{x}$ is continuous on $[0, \infty)$. Now, let $b \in [0, 1]$ be arbitrary. Since $[0, 1] \subset [0, \infty)$, this means that $\forall \epsilon > 0$, $\exists \delta > 0$ s.t. $\mid x-b \mid \implies \mid f(x) - f(b) \mid < \epsilon$ where $f: [0, \infty) \rightarrow \mathbb{R}$ and $f(x) = \sqrt{x}$. Thus, we have shown that $f(x) = \sqrt{x}$ is continuous on $[0, 1]$, as desired.