Why is a real-valued inner product defined on a Euclidean space a continuous function I really need a real maths proof that inner product $\langle \cdot, \cdot \rangle:X \times X \to \mathbb{R}$, where $X$ is an Euclidean space, is a continuous function.
 A: By Cauchy-Schwarz:
$$\lvert\langle x,y \rangle\rvert \leq \lVert x \rVert \cdot \lVert y \rVert$$ 
So $\langle \cdot,\cdot\rangle$ is a bounded bilinear form, and thus continuous.
A: Hint: Note that for any $x,y,\Delta x,\Delta y\in X$, we have $$\begin{align}\langle x+\Delta x,y+\Delta y\rangle &= \langle x,y+\Delta y\rangle+\langle\Delta x,y+\Delta y\rangle\\ &= \langle x,y\rangle+\langle x,\Delta y\rangle+\langle\Delta x,y\rangle+\langle\Delta x,\Delta y\rangle,\end{align}$$ so $$\begin{align}\bigl|\langle x+\Delta x,y+\Delta y\rangle-\langle x,y\rangle\bigr| &\le \bigl|\langle x,\Delta y\rangle\bigr|+\bigl|\langle\Delta x,y\rangle\bigr|+\bigl|\langle\Delta x,\Delta y\rangle\bigr|\\ &\le \lVert x\rVert\lVert\Delta y\rVert+\lVert\Delta x\rVert\lVert y\Vert+\lVert\Delta x\rVert\lVert\Delta y\rVert.\end{align}$$ You must show that for any $\epsilon>0$ and any $x,y\in X$, there is some $\delta>0$ such that $$\bigl|\langle x+\Delta x,y+\Delta y\rangle-\langle x,y\rangle\bigr|<\epsilon$$ whenever $\lVert\Delta x\rVert,\lVert\Delta y\rVert<\delta.$
If you take this approach, you'll want to deal separately with cases that (i) $x,y=0$, (ii) one of $x,y=0$ and the other is not $0$, and (iii) $x,y\neq 0$.

Added: Case (i) is easy. Let me walk you through case (ii), and I'll leave case (iii) to you.
Suppose one of $x,y=0$ and the other is not--without loss of generality, say $x=0,y\neq0$. By the work above, we know that for any $\Delta x,\Delta y\in X$ we have $$\bigl|\langle 0+\Delta x,y+\Delta y\rangle-\langle 0,y\rangle\bigr| \le \lVert 0\rVert\lVert\Delta y\rVert+\lVert\Delta x\rVert\lVert y\Vert+\lVert\Delta x\rVert\lVert\Delta y\rVert=\lVert\Delta x\rVert\lVert y\Vert+\lVert\Delta x\rVert\lVert\Delta y\rVert.$$ Now, since $y\neq 0,$ then $\lVert y\rVert>0$, so $$\lVert\Delta x\rVert\lVert y\rVert<\frac{\epsilon}{2}\qquad\text{if and only if}\qquad\lVert\Delta x\rVert<\frac{\epsilon}{2\lVert y\rVert}.$$ Put $$\delta=\min\left\{\sqrt{\frac{\epsilon}2},\frac{\epsilon}{2\lVert y\rVert}\right\}.$$ Then $\delta>0$, and whenever $\lVert\Delta x\rVert,\lVert\Delta y\rVert<\delta,$ we have $$\begin{align}\bigl|\langle 0+\Delta x,y+\Delta y\rangle-\langle 0,y\rangle\bigr| &\le \lVert\Delta x\rVert\lVert y\Vert+\lVert\Delta x\rVert\lVert\Delta y\rVert\\ &< \frac{\epsilon}{2\lVert y\rVert}\lVert y\rVert+\sqrt{\frac{\epsilon}2}\sqrt{\frac{\epsilon}2}\\ &= \frac{\epsilon}2+\frac{\epsilon}2\\ &= \epsilon,\end{align}$$ as desired.
A: You have an explicit formula for the inner product: $$ \langle x,y \rangle=x_1 y_1+\ldots x_n  y_n $$
This is a polynomial, therefore it is continuous (In fact it is even smooth and analytic)
