Is my proof of the following statement valid? Any feedback is welcome. Thank you in advance.
$f(x) = \sqrt{x}$ is continuous on $[0, \infty)$.
Proof. We break the proof into two cases: (case 1) $a > 0$ and (case 2) $a = 0$.
(case 1) Suppose $a > 0$. Fix $\epsilon > 0$. Let $\delta = \min\{a, \epsilon^2\} > 0$. \begin{align} 0 < |x - a| < \delta &\implies a - \delta < x < a + \delta \\ &\implies \sqrt{a - \delta} < \sqrt{x} < \sqrt{a + \delta} \qquad \text{(by Lemma 1 below and $(a - \delta) \geq 0 $ because $\delta = \min\{a, \epsilon^2\}$)} \\ &\implies \sqrt{a} - \sqrt{\delta} < \sqrt{x} < \sqrt{a} + \sqrt{\delta} \qquad \text{(by Lemma 2 below)} \\ &\implies -\sqrt{\delta} < \sqrt{x} - \sqrt{a} < \sqrt{\delta} \\ &\implies |\sqrt{x} - \sqrt{a}| < \sqrt{\delta} \leq \sqrt{\epsilon^2} \\ &\implies |\sqrt{x} - \sqrt{a}| < \epsilon \\ &\implies |f(x) - f(a)| < \epsilon. \end{align} This proves that $f$ is continuous on $(0, \infty)$.
(case 2) Suppose $a = 0$. Fix $\epsilon > 0$. Let $\delta = \epsilon^2 > 0$. \begin{align} 0 < |x - a| < \delta &\implies 0 < |x| < \delta \\ &\implies 0 < x < \delta \\ &\implies \sqrt{x} < \sqrt{\delta} = \sqrt{\epsilon^2} \qquad \text{(by Lemma 1 below)}\\ &\implies \sqrt{x} < \epsilon \\ &\implies |\sqrt{x} - 0| < \epsilon \\ &\implies |f(x) - f(a)| < \epsilon. \end{align} This proves that $f$ is continuous at $0$.
Therefore, $f$ is continuous on $[0, \infty)$. $\square$
Lemma 1. If $0 \leq a < b$, then $\sqrt{a} < \sqrt{b}$.
Proof. (Lemma 1) We will prove by contradiction. Suppose $0 \leq a < b$ and $\sqrt{a} \geq \sqrt{b}$. Since $\sqrt{a}, \sqrt{b} \geq 0$ by definition, multiplying both sides of $\sqrt{a} \geq \sqrt{b}$ by $\sqrt{a}$ and $\sqrt{b}$ respectively gives us the following. $$ (a \geq \sqrt{a} \sqrt{b} \quad \text{and} \quad \sqrt{a} \sqrt{b} \geq b) \implies a \geq b \quad \textbf{contradiction!} \square $$
Lemma 2. If $a, b > 0$, then $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$.
Proof. (Lemma 2) By the triangle inequality we have the following. \begin{align} |\sqrt{a} + \sqrt{b}| \leq |\sqrt{a}| + |\sqrt{b}| &\implies \sqrt{(\sqrt{a} + \sqrt{b})^2} \leq \sqrt{(\sqrt{a})^2} + \sqrt{(\sqrt{b})^2} \\ &\implies \sqrt{a + b} \leq \sqrt{a + 2\sqrt{a}\sqrt{b} + b} \leq \sqrt{a} + \sqrt{b} \\ &\implies \sqrt{a + b} \leq \sqrt{a} + \sqrt{b}. \square \end{align}