For finite dimensional spaces, the answer is "yes"; this is a consequence of the Gram-Schmidt orthonormalization process: every finite dimensional inner product space over $\mathbb{R}$ or over $\mathbb{C}$ has an orthonormal basis.
Now let $\mathbf{V}$ be an inner product space, and let $\mathbf{v}_1,\ldots,\mathbf{v}_n$ be an orthonormal basis. Then $T\colon\mathbf{V}\to \mathbf{F}^n$ given by $T\mathbf{v}_i = \mathbf{e}_i$ (i.e., $T$ maps each vector in $\mathbf{V}$ to its coordinate vector relative to the orthonormal basis $\mathbf{v}_1,\ldots,\mathbf{v}_n$) is an invertible linear transformation such that for all $\mathbf{x},\mathbf{y}\in\mathbf{V}$, $\langle \mathbf{x},\mathbf{y}\rangle = T\mathbf{x}\cdot T\mathbf{y}$, where the right hand side is the standard dot product on $\mathbf{F}^n$ ($\mathbf{F}=\mathbb{R}$ or $\mathbb{C}$).
I find it cleanest to think in terms of Lipschitzness, rather than the $\varepsilon$-$\delta$ definition of continuity.
Let $(X, d)$ be a metric space.
For any $a \in X$, the map $d (a, \cdot) : X \to \mathbb R$ is Lipschitz, and hence is continuous.
The map $d : X \times X \to \mathbb R$ is Lipschitz w.r.t. the product metric, and hence is continuous.
I will prove (2.) and leave (1.) as a simpler exercise. (In fact, (2.) also directly implies (1.).) Fix $(x, y), (z, w) \in X \times X$. Then
$$
\begin{array}{rll}
|d(x,y) - d(z,w)|
&\leqslant |d(x,y) - d(z,y)| + |d(z,y) - d(z,w)| &
\\ &\leqslant d(x, z) + d(y, w) & \text{(triangle inequality on } d \text{)}
\\ &\leqslant 2 \max \{ d(x, z) , d(y, w) \} &
\\ &= 2 d_{\infty} ((x, y) , (z, w)), &
\end{array}
$$
showing that $d$ is Lipschitz. (Here we have assumed the “max” metric on the product; certainly other choices are possible but they turn out to be equivalent.)
Coming to the OP's question, to prove that a norm on a linear space is continuous w.r.t. itself, notice that the linear space is also a metric space with $d(x,y) = \| x - y \|$. In other words, $\| x \|$ is just the distance of $x$ from the origin; hence applying item (1.) above (with $a = 0$) shows the continuity of the norm.
A point to ponder: Equivalent norms. The above discussion might suggest that continuity of a norm is not an interesting thing to study. However this question can be modified a little to give rise to a quite fruitful concept:
Given two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear space $X$, are they continuous with respect to each other? I.e., when $\| \cdot \|_1$ is viewed as a function, is it continuous w.r.t. the norm $\| \cdot \|_2$ and vice versa?
It can be shown that this is true if and only if there exist numbers $0 < \alpha \leqslant \beta < \infty$ such that
$$
\alpha \| v \|_1 \leqslant \| v \|_2 \leqslant \beta \| v \|_1
$$
for all vectors $v$. In this case, we say that the two norms are equivalent.
Best Answer
Yes.
Fix x in the inner product space, and let $f(y) = \langle y, x \rangle$ denote the inner product function. Note that this is a linear functional -- that is, it is linear in y, and maps vectors to scalars.
It is a well-known theorem that linear functionals are continuous (on the entire space) if and only if they are bounded. Here, "bounded" means that there exists a constant M such that $|f(y)| \leq M|y|$ for all y in the space.
That the inner product functional is bounded now follows from the Cauchy-Schwarz Inequality: $|f(y)| \leq |x||y|.$