Is Inner product continuous when one arg is fixed?

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: $|f(y)| \leq |x||y|.$

Fix $x$. We have $|\langle x,y\rangle - \langle x,z\rangle| = |\langle x,y-z\rangle|\leq ||x||||y-z||$ by the Cauchy-Schwarz inequality. This gives you easily that $\langle x,-\rangle\colon \mathbf{V}\to\mathbb{F}$ is not only continuous, but uniformly continuous. Similarly for $\langle -,x\rangle$.

If we define $T_x v = \langle v,x\rangle$ then $T_x$ is a linear functional. Since an inner product induces a norm, and thus it is continuous if and only if it is bounded on the unit circle.

And so $|T_x v | = |\langle v,x\rangle | \le \|v\|\cdot \|x\|$ and for $v$ on the unit circle, i.e. $\|v\| = 1$ we have that $T_x$ is indeed bounded and therefore continuous.

The proof is similar if you fix the left argument, however it may not be a linear functional if it is a complex vector space, but rather antilinear (that is linear up to conjugation). The norm is unaffected by that, though, so it's not a big step to overcome.