Suppose $X$ is open in $\mathbb{R}^{n}$ and $F$ is a Banach space. If $f: X \rightarrow F$ is continuously differentiable, then $f$ has continuous partial derivatives.
Could you please verify whether my attempt on this well-known result is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb{R}^{n}$.
It follows from the existence of $\partial f(\cdot):X \to \mathcal L(\mathbb R^n,F)$ that $\partial_k f(\cdot) = \partial f(\cdot)(e_k)$ and thus $$\begin{aligned}\|\partial_k f(x) -\partial_k f(a)\| &=\|\partial f(x)(e_k) - \partial f(a)(e_k)\|\\ &= \|(\partial f(x)-\partial f(a))(e_k)\| \\ &\le \|\partial f(x)-\partial f(a)\| \cdot \|e_k\|\\ &= \|\partial f(x)-\partial f(a)\|\end{aligned}$$
Because $\partial f(\cdot):X \to \mathcal L(\mathbb R^n,F)$ is continuous, $\|\partial f(x)-\partial f(a)\| \to 0$ and thus $\|\partial_k f(x) -\partial_k f(a)\| \to 0$ as $x \to a$. Hence $\partial_k f (\cdot): X \to \mathcal L(\mathbb R,F)$ is continuous.