The Definition of Continuity

Constants $c$: $\lvert c - c\rvert = 0 < \varepsilon$, so continuous. The identity $x \mapsto x$: $\delta = \varepsilon$ works, since $\lvert x - a\rvert < \delta$ gives $\lvert x - a\rvert < \varepsilon$.

(Sum) Let $f, g$ be continuous at $a$. Given $\varepsilon > 0$, take $\delta_1$ with $\lvert f(x) - f(a)\rvert < \frac{\varepsilon}{2}$ and $\delta_2$ with $\lvert g(x) - g(a)\rvert < \frac{\varepsilon}{2}$; with $\delta = \min(\delta_1, \delta_2)$, $\lvert (f+g)(x) - (f+g)(a)\rvert \leq \lvert f(x) - f(a)\rvert + \lvert g(x) - g(a)\rvert < \varepsilon$.

(Product) Use the identity $\lvert f(x)g(x) - f(a)g(a)\rvert \leq \lvert f(x)\rvert\cdot\lvert g(x) - g(a)\rvert + \lvert g(a)\rvert\cdot\lvert f(x) - f(a)\rvert$.

First bound $f$ near $a$: by continuity of $f$ there is $\delta_0$ with $\lvert f(x) - f(a)\rvert < 1$ for $\lvert x - a\rvert < \delta_0$, hence $\lvert f(x)\rvert \leq \lvert f(a)\rvert + 1 =: M$ there (local boundedness).

Now take $\delta_1$ with $\lvert g(x) - g(a)\rvert < \frac{\varepsilon}{2M}$ and $\delta_2$ with $\lvert f(x) - f(a)\rvert < \frac{\varepsilon}{2(\lvert g(a)\rvert + 1)}$; with $\delta = \min(\delta_0, \delta_1, \delta_2)$ we get $\lvert f(x)g(x) - f(a)g(a)\rvert < M \cdot \frac{\varepsilon}{2M} + \lvert g(a)\rvert \cdot \frac{\varepsilon}{2(\lvert g(a)\rvert+1)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$.

We show $x^n$ is continuous by induction.

Base cases: for $n = 0$, $x^0 = 1$ is the constant function proved continuous above; for $n = 1$, $x^1 = x$ is the identity proved continuous above.

Inductive step: if $x^n$ is continuous, then $x^{n+1} = x^n \cdot x$ is a product of continuous functions, hence continuous by the product rule.

So $x^n$ is continuous for every $n \ge 0$.

Then $c x^n$ is continuous (constant times continuous, by the product rule), and any polynomial — a finite sum of such terms $c_k x^k$ — is continuous (the sum rule, applied finitely many times).

We need $\lvert x - 0\rvert < \delta \Rightarrow \lvert \lvert x\rvert - \lvert 0\rvert \rvert < \varepsilon$. But $\lvert \lvert x\rvert - 0\rvert = \lvert x\rvert$, so it suffices to get $\lvert x\rvert < \varepsilon$.

Put $\delta = \varepsilon$: then $\lvert x - 0\rvert = \lvert x\rvert < \delta = \varepsilon$ is exactly the desired inequality. Hence $\lvert x\rvert$ is continuous at $0$.

[framing] Continuity does not imply differentiability: $\lvert x\rvert$ is not differentiable at $0$. Continuity is "no breaks", not "smoothness".

Let $\varepsilon > 0$. Since $g$ is continuous at $b = f(a)$, there is $\eta > 0$ with $\lvert g(y) - g(b)\rvert < \varepsilon$ whenever $\lvert y - b\rvert < \eta$.

Now apply the continuity of $f$ to $\eta$: there is $\delta > 0$ with $\lvert f(x) - f(a)\rvert < \eta$ whenever $\lvert x - a\rvert < \delta$.

If $\lvert x - a\rvert < \delta$ then $\lvert f(x) - b\rvert < \eta$, so $\lvert g(f(x)) - g(b)\rvert < \varepsilon$, i.e. $\lvert h(x) - h(a)\rvert < \varepsilon$. Hence $h$ is continuous at $a$.

Numerically: if $g$ demands $\eta = \frac{\varepsilon}{2}$ and $f$ answers that $\eta$ with $\delta = \frac{\eta}{2}$, the composition runs on $\delta = \frac{\varepsilon}{4}$ — the allowances compose along the chain.