The inner product is jointly continuous

Inserting and subtracting the mixed pairing gives $⟨x,y⟩-⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩=⟨x-x^{\mathrm{\prime }},y⟩+⟨x^{\mathrm{\prime }},y-y^{\mathrm{\prime }}⟩$, so Cauchy–Schwarz applied to the two summands and the triangle inequality for scalars give $\mathrm{\mid }⟨x,y⟩-⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩\mathrm{\mid }\le \mathrm{\parallel }x-x^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }+\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }$.

Given $\epsilon >0$, put $\delta =\min \{1,\epsilon \mathrm{/}(1+\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel })\}>0$; if $\mathrm{\parallel }x-x^{\mathrm{\prime }}\mathrm{\parallel }<\delta$ and $\mathrm{\parallel }y-y^{\mathrm{\prime }}\mathrm{\parallel }<\delta$, then $\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel }+\delta <1+\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel }$ by the triangle inequality, so step 1.1 gives $\mathrm{\mid }⟨x,y⟩-⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩\mathrm{\mid }<\delta (1+\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel })\le \epsilon$.

The product $B(x^{\mathrm{\prime }},\delta )\times B(y^{\mathrm{\prime }},\delta )$ is a basic product-open set containing $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$, and step 2.1 shows that on it the pairing stays within every ball about $⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩$, so the pairing is continuous at every point of $V\times V$; if moreover $x_n\to x$ and $y_n\to y$, then for every $\epsilon >0$ the pair $(x_n,y_n)$ eventually lies in the corresponding $\delta$-box, whence $\mathrm{\mid }⟨x_n,y_n⟩-⟨x,y⟩\mathrm{\mid }<\epsilon$ and $⟨x_n,y_n⟩\to ⟨x,y⟩$.