Prove that continuous partial derivatives imply continuous total derivative Good morning, I'm trying to prove that

Suppose $X$ is open in $\mathbb{R}^{n}$ and $F$ is a Banach space. Then $f: X \rightarrow F$ is continuously differentiable if $f$ has continuous partial derivatives.

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:
For $a \in X$, we define $A \in \mathcal L(\mathbb R^n,F)$ by $$h=\left(h_{1}, \ldots, h_{n}\right) \mapsto A h=\sum_{k=1}^{n} \partial_{k} f(a) h_{k}$$
Our goal is to show that $\partial f(a) = A$ or equivalently $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$
First, we choose $\varepsilon>0$ such that $\mathbb{B}(a, \varepsilon) \subseteq X$ and let $x_k = a+ (h_1,\ldots,h_k,0,\ldots,0)$ for all $k = \overline{1,n}$. It follows that $$f(a+h)-f(a)=\sum_{k=1}^{n}\left(f\left(x_{k}\right)-f\left(x_{k-1}\right)\right)$$
Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$. By definition, we have $$\begin{aligned} 
\partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) &=  \lim_{z \to 0} \frac{f\left(x_{k-1}+t h_{k} e_{k} + ze_k\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z} \\ 
 &=  \lim_{z \to 0} \frac{f\left(x_{k-1}+ (th_{k}+z)  e_{k}\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z}\\
 &=  \frac{\partial f\left(x_{k-1}+t h_{k} e_{k}\right)}{\partial (th_k)}   
\end{aligned}$$
By Fundamental Theorem of Calculus, we have $$\begin{aligned} 
h_k\int_0^1 \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) dt &= \int_0^1 \partial_{k} f\left(x_{k-1}+t h^{k} e_{k}\right) d(th_k)\\ &=  \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) \Big|_0^1 \\
&= f\left(x_{k}\right)-f\left(x_{k-1}\right)
\end{aligned}$$
As such, $$f(a+h)-f(a)=\sum_{k=1}^{n} h_{k} \int_{0}^{1} \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) d t$$
Consequently, $$\begin{aligned} \|f(a+h)-f(a) - Ah \| &=\left \|\sum_{k=1}^{n} h_{k} \int_{0}^{1} \left(\partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a)\right) d t \right \| \\ &\le \sum_{k=1}^{n} |h_{k}|  \int_{0}^{1} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a)\right \| d t \\ &\le |h|_\infty \sum_{k=1}^{n} \int_{0}^{1} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a)\right \| d t \\ &\le |h|_\infty \sum_{k=1}^{n}   \int_{0}^{1} \sup_{t \in [0,1]} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a) \right \| d t \\ &\le |h|_\infty \sum_{k=1}^{n}   \int_{0}^{1}  \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| d t \\&= |h|_\infty \sum_{k=1}^{n}  \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \|\end{aligned}$$
We have $h \to 0$ implies $\|h\|_\infty \to 0$, which in turn implies $x \to a$. It follows from the continuity of $\partial_{k} f\left(x\right)$ that $ \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| \to 0$ as $x \to a$.
Finally,  $$\lim _{h \rightarrow 0} \frac{\| f(a+h)-f(a)-A h \|}{|h|_\infty} \le \lim _{h \rightarrow 0} \sum_{k=1}^{n}  \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| = 0$$ Consequently, $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$
Hence $\partial f(a) = A$. Next we prove that $\partial f(\cdot): X \to \mathcal L(\mathbb R^n,F)$ is continuous. We have $$\begin{aligned}\|\partial f(x)h - \partial f(a)h\| &= \left\| \sum_{k=1}^{n} \partial_{k} f\left(x\right) h_{k} - \sum_{k=1}^{n} \partial_{k} f\left(a\right) h_{k} \right\| \\ &= \left\| h_k \sum_{k=1}^{n} ( \partial_{k} f\left(x\right) - \partial_{k} f\left(a\right)) \right\| \\&\le  \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \cdot |h|_\infty \end{aligned}$$
Consequently, $$\|\partial f(x) - \partial f(y)\| = \sup_{h \in X} \frac{\|\partial f(x)h - \partial f(a)h\|}{|h|_\infty} \le \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\|$$
It follows from the continuity of $\partial_{k} f\left(\cdot\right)$ that $\sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \to 0$ and thus $\|\partial f(x) - \partial f(y)\| \to 0$ as $x \to a$. Hence $\partial f(x) \to \partial f(y)$.
 A: I've just figured out a variant that utilized the MVT, so I posted it here. I would be great if someone helps me verify it. Thank you so much for your help!

My attempt:
For $a \in X$, we define $A \in \mathcal L(\mathbb R^n,F)$ by $ A h=\sum_{k=1}^{n} \partial_{k} f(a) h_{k}$ for all $h=\left(h_{1}, \ldots, h_{n}\right) \in X$. Our goal is to show that $\partial f(a) = A$ or equivalently $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$
Let $x_k = a+ (h_1,\ldots,h_k,0,\ldots,0)$ for all $k = \overline{1,n}$. It follows that $f(a+h)-f(a)=\sum_{k=1}^{n}\left(f\left(x_{k}\right)-f\left(x_{k-1}\right)\right)$. Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$. By definition, we have $$\begin{aligned} 
h_k\partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) &= h_k  \lim_{z \to 0} \frac{f\left(x_{k-1}+t h_{k} e_{k} + ze_k\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z} \\ 
 &=  \lim_{z \to 0} \frac{f\left(x_{k-1}+ th_{k}e_k+(z/h_k)h_ke_{k}\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z/h_k}\\
 &=  \lim_{z' \to 0} \frac{f\left(x_{k-1}+ th_{k}e_k+z'h_ke_{k}\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z'}\\
 &=  \lim_{z \to 0} \frac{f\left(x_{k-1}+ (t+z')h_{k}e_k\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z'}\\
 &=  \frac{\partial f\left(x_{k-1}+t h_{k} e_{k}\right)}{\partial t}   
\end{aligned}$$
We apply Mean Value Theorem for the map $\mathbb R \to F, \quad t \mapsto f\left(x_{k-1}+t h_{k} e_{k}\right)$ and get $$\begin{aligned}f(x_k) - f(x_{k-1}) &=  f(x_{k-1}+1\cdot h_ke_k)-f(x_{k-1}+0\cdot h_ke_k)\\ &= \frac{\partial f\left(x_{k-1}+t h_{k} e_{k}\right)}{\partial t} (t_k) \\ &= h_k \partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right)\end{aligned}$$
Consequently,
$$\begin{aligned} \|f(a+h)-f(a) - Ah \|
&=\left \|\sum_{k=1}^{n} h_{k} \left(\partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right) -\partial_{k} f(a)\right) \right \| \\
&\le \sum_{k=1}^{n} |h_{k}| \left \| \partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right) -\partial_{k} f(a)\right \| \\
&\le |h|_\infty \sum_{k=1}^{n} \left \| \partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right) -\partial_{k} f(a)\right \| \\
&\le |h|_\infty \sum_{k=1}^{n} \sup_{t \in [0,1]} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a) \right \| \\
&\le |h|_\infty \sum_{k=1}^{n} \sup_{|x-a|_\infty \le |h|_\infty} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \|
\end{aligned}$$
We have $h \to 0$ implies $|h|_\infty \to 0$, which in turn implies $x \to a$. It follows from the continuity of $\partial_{k} f\left(\cdot\right)$ that $$\sup_{|x-a|_\infty \le |h|_\infty} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| \to 0 \quad (x \to a)$$
Finally,  $$\lim _{h \rightarrow 0} \frac{\| f(a+h)-f(a)-A h \|}{|h|_\infty} \le \lim _{h \rightarrow 0} \sum_{k=1}^{n}  \sup_{|x-a|_\infty \le |h|_\infty} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| = 0$$ Consequently, $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$
Hence $\partial f(a) = A$. Next we prove that $\partial f(\cdot): X \to \mathcal L(\mathbb R^n,F)$ is continuous. We have $$\begin{aligned}\|\partial f(x)h - \partial f(a)h\| &= \left\| \sum_{k=1}^{n} \partial_{k} f\left(x\right) h_{k} - \sum_{k=1}^{n} \partial_{k} f\left(a\right) h_{k} \right\| \\ &= \left\| h_k \sum_{k=1}^{n} ( \partial_{k} f\left(x\right) - \partial_{k} f\left(a\right)) \right\| \\&\le |h|_\infty \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \end{aligned}$$
Consequently, $$\|\partial f(x) - \partial f(y)\| = \sup_{h \in X} \frac{\|\partial f(x)h - \partial f(a)h\|}{|h|_\infty} \le \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\|$$
It follows from the continuity of $\partial_{k} f\left(\cdot\right)$ that $\sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \to 0$ and thus $\|\partial f(x) - \partial f(y)\| \to 0$ as $x \to a$. Hence $\partial f(x) \to \partial f(y)$.
