Continuity proof. I want to prove that $\exp x$ and $\sin x$ are continuous. This means I want to show that
$$\lim\limits_{x\to a}e^x=e^a$$
$$\lim\limits_{x\to a}\sin x=\sin a$$
for any fixed $a \in \Bbb R$. Then I need to show that for any $\epsilon >0$ there exists a $\delta >0$ such that whenever $|x-a|<\delta$, then $|\sin x -\sin a|<\epsilon$ and similarily $|e^x-e^a|<\epsilon$. For the first case I have that
$$\left| {\sin x - \sin a} \right| = 2\left| {\sin \frac{{x - a}}{2}} \right|\left| {\cos \frac{{x + a}}{2}} \right|$$
for the second case I use
$$\left| {{e^x} - {e^a}} \right| = \left| {{e^a}} \right|\left| {{e^{x - a}} - 1} \right|$$
ADD Just for the sake of a simple solution:
Let $\epsilon>0$ and $a\in \bf R$ be given. Choose $\delta = \epsilon$. Then we have that 
$$\eqalign{
  \left| {\sin x - \sin a} \right| = 2\left| {\sin \left( {\frac{{x - a}}{2}} \right)\cos \left( {\frac{{x + a}}{2}} \right)} \right| &\cr 
   = 2\sin \left| {\frac{{x - a}}{2}} \right|\cos \left| {\frac{{x + a}}{2}} \right| &\cr 
   \leqslant 2\left| {\frac{{x - a}}{2}} \right| = \left| {x - a} \right| < \delta  = \epsilon  &\cr} $$
For the exponential, we have that $\log:(0,+\infty)\to \Bbb R$ is continuous, one-one, onto, and differentiable on $(0,+\infty)$. It follows that $\exp:\mathbb R\to (0,\infty)$ defined by $\exp x =y \iff \log y = x$ is also continuous and differentiable.
Definitions:
$y=e^x$ is the inverse of $y=\log x $ defined as
$$\log x = \lim_{k \to 0} \frac{x^k-1}{k}$$
Here I gave a list of the elementary functions of $\log$.
I would probably choose to define $\sin$ like this 
Checking Apostol's Calculus I found he proves the continuity of $\sin x$ by first proving that if $f$ is (Riemann) integrable in $[a,x]$ for all $x\in [a,b]$ then 
$$F(x) = \int_a^x f(t) dt$$
is continuous for all $x\in [a,b]$. Under this light it is evident
$$\sin x = \int_0^x \cos t dt $$
$$\cos x = 1-\int_0^x \sin t dt $$
$$e^x = \int_0^x e^t dt+1$$
are all continuous.
 A: There's a sense of deja vu here. I'm not exactly what you consider axiomatic.
Here's a different approach for $\exp$.
The function $x \mapsto \exp(x)$ is, by your definition, the inverse of $x \mapsto \log(x)$. The function $\log$ is continuous, strictly increasing, and the range is $\mathbb{R}$. Hence it follows that $\exp$ is strictly increasing and defined on all of $\mathbb{R}$. Let $\epsilon>0$, and let $\delta = \min\{-\log(1-\epsilon), \log(1+\epsilon)\}$, and suppose $|x| < \delta$. Then since
$$\log(1-\epsilon) < x < \log(1+\epsilon)$$
we have
$$ 1-\epsilon = \exp(\log(1-\epsilon)) < \exp(x) < \exp(\log(1+\epsilon)) = 1+\epsilon$$
Hence $|\exp(x)-1| < \epsilon$, and so $\exp$ is continuous at $0$.
Original 'non-answer':
For $\sin$, you know that $\sin^2(x)+\cos^2(x) = 1$, hence $\cos(x)$ is bounded by $1$. Since $\cos$ is the derivative of $\sin$, the mean value theorem shows that $|\sin(x)-\sin(y)| \leq |x-y|$ for any $x,y$.
Replacement answer:
In context of the definition of $\sin$ from the question: The function $x \mapsto A(x)$ is strictly increasing and continuous on $[-1,1]$. Using a similar argument to that used for $\exp$ above, the function $\cos$ can be seen to be continuous. Hence, since $\sin(x) = \sqrt{1-\cos^2(x)}$, it follows that $\sin$ is continuous.
A: For the case of continuity of SINX, after what you made,  I would use that absolute value of SIN(u(x))
is lower or equal than absolute value of U(X), and absolute value of cos(U(X)) is lowre or equal to 1.
A: Proof that both $sin(x)$ and $cos(x)$ are uniformly continuous on $[0, 2 \pi]$.
Let $I = [0, 2 \pi]$ be subdivided into $2^n$ subintervals, as follows
$x_{k} = k \frac{2 \pi}{2^n}\;$ for $0 \le k \le 2^n$.
Define $z_0 = 1$, $z_1 = -1$, $z_2 = i$, and 
$z_{m+1} = \sqrt {z_m}$, with $Re(z_{m+1}) > 0$ for $m \ge 2$
For each $n$ define the discrete function $W_n$ as
$k \frac{2 \pi}{2^n}\; \mapsto z_n^k$,  $\,0 \le k \le 2^n$.
sending a finite number of points into the complex circle.
Define the $cos$ and $sin$ functions as follows:
$z_n^k = cos(k \frac{2 \pi}{2^n}) + sin(k \frac{2 \pi}{2^n})i$
The graphs of the functions $W_n$ satisfy $W_n \subset W_{n+1}$, so we can take the union of these sets and get a new function $W$ defined on a dense subset $D$ of $I$.
Proposition 1: The function $W$ is uniformly continuous on $D$, taking countably infinite values on the unit circle. Both $sin(x)$ and $cos(x)$ are also uniformly continuous.
Proof: Compare the lengths of chords and arcs on the unit circle.
Proposition 2: The function $W$ is uniformly continuous on $I$, with range the unit circle. Both $sin(x)$ and $cos(x)$ are also uniformly continuous.
Proof: Extend by continuity.
