$(\delta,\epsilon)$ Continuity Proof $$
f(T) : \mathbb{R}^2 \to \mathbb{R}^2\\
f(T) = \begin{pmatrix}  \sin(x)\sin(y) \\y\end{pmatrix}
$$
I have to prove that f is a continuous function in $\mathbb{R}^2$ using $\epsilon$ and $\delta$
My try ended up in a problem 
for all $\epsilon>0$ Exist $\delta>0$ so that $\lVert T-a\rVert < \delta =\epsilon$ for all $a\in \mathbb{R}^2$ 
to prove $\lVert f(T) -f(a)\rVert < \epsilon$
$$
\lVert f(T) -f(a)\rVert= \sqrt{\bigl[(\sin(x)\sin(y)-\sin(a_1)\sin(a_2)\bigr]^2+(y-a_2)^2}
$$
how can I prove  that  $\bigl[(\sin(x)\sin(y)-\sin(a_1)\sin(a_2)\bigr]^2 <(x-a_1)^2$ ? 
or  did i have a wrong  start ? 

$$|\sin x\sin y-\sin a_1\sin a_2|\\
=|\sin x\sin y-\sin a_1\sin y+\sin a_1\sin y-\sin a_1\sin a_2|\\
\le |\sin x-\sin a_1|\cdot|\sin y|+|\sin a_1|\cdot|\sin y-\sin a_2|\\
\le |x-a_1|+|y-a_2|
$$
So:
$$(\sin x\sin y-\sin a_1\sin a_2)^2\le (|x-a_1|+|y-a_2|)^2\le 2(x-a_1)^2+2(y-a_2)^2$$
I used $|\sin\alpha-\sin\beta|\le|\alpha-\beta|$ (simple application of Mean Value Theorem), and $(\alpha+\beta)^2\le 2\alpha^2+2\beta^2$
Then you can say:
$\lVert f(T)-f(a)\rVert\le\sqrt{2(x-a_1)^2+3(y-a_2)^2}\le\sqrt{3(x-a_1)^2+3(y-a_2)^2}=\sqrt{3}\lVert T-a\rVert$
so you may take $\delta=\frac{\epsilon}{\sqrt 3}$
