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: $|f(y)| \leq |x||y|.$
We will use the infinity or max norm on $V\times V$, that is, for $(u,v)\in V\times V$, $\lVert (u,v)\rVert_{V\times V}=\max\{\lVert u\rVert_V,\lVert v\rVert_V\}$, where the norm on $V$ is the norm induced by the inner product $\langle\cdot,\cdot\rangle$. I can't speak with authority on why this is equivalent to the product topology, but I don't think it should be too difficult to show for a finite product of normed spaces - those are some well-behaved constructs. For a function to be jointly continuous, it means simply that it is continuous in the product topology, rather than only being continuous in each variable separately, with the others fixed. That is, for $f:X\times Y\to Z$, $f$ is jointly continuous if, for all $(x_0,y_0)\in X\times Y$, given any $\epsilon>0$, there exists a $\delta>0$ such that, for all $(x,y)\in X\times Y$, $d_{X\times Y}((x,y),(x_0,y_0))<\delta$ implies $d_Z(f(x,y),f(x_0,y_0))<\epsilon$. So let's do it, using the induced norm as our metric.
Fix an $x_0,y_0\in V$, and let $\epsilon>0$ be given. Choose $\delta=\min\{1,\frac{\epsilon}{2(\lVert y_0\rVert + 1)},\frac{\epsilon}{2(\lVert x_0\rVert+1)}\}$, and note $\lvert\lVert y\rVert_V-\lVert y_0\rVert_V\rvert\leq \lVert y-y_0\rVert_V<1$, by the reverse triangle inequality and our choice of $\delta$. This implies $\lVert y\rVert_V<1+\lVert y_0\rVert_V$.
Let $x,y\in V$, and suppose $\lVert (x,y)-(x_0,y_0)\rVert_{V\times V}=\lVert(x-x_0,y-y_0)\rVert_{V\times V}<\delta$, implying $\lVert x-x_0\rVert_V<\delta$ and $\lVert y-y_0\rVert_V<\delta$. Then
$\lvert\langle x,y\rangle-\langle x_0,y_0\rangle\rvert = \lvert\langle x,y\rangle-\langle x_0,y\rangle+\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$
$\leq \lvert\langle x,y\rangle-\langle x_0,y\rangle\rvert+\lvert\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$
$= \lvert\langle x-x_0,y\rangle\rvert+\lvert\langle x_0,y-y_0\rangle\rvert$
$\leq \lVert x-x_0\rVert_V\lVert y\rVert_V + \lVert x_0\rVert_V \lVert y-y_0\rVert_V$ (by Cauchy-Bunyakovsky-Schwarz inequality)
$< (\frac{\epsilon}{2(\lVert y_0\rVert+1)})(1+\lVert y_0\rVert_V) + (\lVert x_0\rVert_V+1)(\frac{\epsilon}{2(\lVert x_0\rVert+1)})$, (since $\lVert x_0\rVert_V<\lVert x_0\rVert_V+1$, and by our choice of $\delta$)
$= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$.
Best Answer
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:
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.