A proof on Banach space with two variables We consider functions $f:\mathbb R^2 \mapsto \mathbb R$ with two functionals defined by
$$|f|_0 = \sup_{x, y \in \mathbb R} |f(x,y)|$$
and
$$|f|_{0,1} = |f|_0 + |\partial_x f|_0,$$
whenever they exist.
In the above, $\partial_x$ means the partial derivative to the first variable, similarly $\partial_y$ means the partial derivative to the second variable.
Define the space $S = \{f: |f|_{0,1}<\infty\}$. One can check $|\cdot|_{0,1}$ is a norm in $S$. In particular, triangle inequality holds due to
$$
|f+ g|_{0,1} = 
|f+g|_0 + |\partial_x(f+g)|_{0,1}
\le 
|f|_{0,1} + |g|_{0,1}.$$
Q. Is $S$ a Banach space?
It seems to me the following two things are valid:
For any Cauchy sequence $\{f_n\}$ in $S$, they are also Cauchy in $|\cdot|_0$, so there exists $f$ with
$$|f_n- f|_0 \to 0.$$
Similarly, there exists $g$ such that
$$|\partial_x f_n - g|_0 \to 0.$$
However, one may need to prove that $\partial_x f = g$ for the last piece.
 A: $\newcommand{\md}[1]{{\left\lvert #1 \right\lvert}}$
Let us prove the following result.

Theorem. Suppose $\{f_n\}$ is a sequence of real functions, with first partial derivative existing on $\mathbb{R}^2$ such that $\{f_n\}$ and $\{\partial_x f_n\}$ both converge uniformly. Let $f$ denote the limit of $\{f_n\}$. Then, $$\partial_x f(a, b) = \lim_{n\to \infty} \partial_x f(a, b) \tag{1}$$
for all $(a, b) \in \mathbb{R}^2$.
(Implicit in the above equality is the fact that the partial derivative on the left does exist.)

Proof. Fix $(a, b) \in \mathbb{R}^2$ and $\varepsilon > 0$.
By uniform convergence of $\{\partial_x f_n\}$, there exists $N \in \mathbb{N}$ such that
$$\md{\partial_x f_n(u, b) - \partial_x f_m(u, b)} < \varepsilon \tag{2}$$
for all $u \in \mathbb{R}$ and all $m, n \geqslant N$.
Applying the (appropriate version of the) mean value theorem to $f_n - f_m$, $(2)$ gives us
$$\md{f_n(t, b) - f_m(t, b) - f_n(a, b) + f_m(a, b)} \leqslant \md{t - a}\varepsilon \tag{3}$$
for all $t \in \mathbb{R}$ and $m, n \geqslant N$.
Now, define $\varphi_n$ and $\varphi$ on $\dot I := \mathbb{R} \setminus \{a\}$ by
$$\varphi_n(t) := \frac{f_n(t, b) - f_n(a, b)}{t - a} \quad\text{and}\quad \varphi(t) := \frac{f(t, b) - f(a, b)}{t - a}.$$
Then, for all $n \geqslant 1$, we have
$$\lim_{t\to a} \varphi_n(t) = \partial_x f_n(a, b). \tag{4}$$
By $(3)$, we have
$$\md{\varphi_n(t) - \varphi_m(t)} \leqslant \varepsilon$$
for all $t \in \dot I$ and $m, n \geqslant N$. Thus, $\{\varphi_n\}$ converges uniformly. Moreover, since $f_n \to f$, we get
$$\varphi_n \xrightarrow{\text{uniformly}} \varphi. \tag{5}$$
Since the convergence is uniform, we can interchange limits. That is, $(4)$ and $(5)$ show us that
$$\lim_{t\to a} \varphi(t) = \lim_{n\to \infty} \partial_x f_n(a, b).$$
The above is precisely $(1)$. $\qquad \Box$

From this, your result follows. Indeed, $\{f_n\}$ be a Cauchy sequence in the $\md{\cdot}_{0, 1}$ norm on $S$. Then, $\{f_n\}$ and $\{\partial_x f_n\}$ are Cauchy in the $\md{\cdot}_{0}$ norm on $S$. (This is also indeed a norm on $S$.)
Thus, both the sequences have a limit, and the convergence to the limit is also uniform. The theorem now applies readily.
