I want to prove the following result
Proposition 1
Let $T \in D'(\mathbb R^n)$ and $S \in D'(\mathbb R^m)$. Then $$ \langle T_x, \langle S_y, \varphi(x,y) \rangle \rangle= \langle S_y \langle T_x, \varphi(x,y) \rangle \rangle \quad \forall \varphi \in C_0^\infty(\mathbb R^n \times \mathbb R^m ),$$
using the fact:
The set $S$ of all linear combinations of tensor products of functions in $C_0^\infty(\mathbb R^n)$ and $C_0^\infty(\mathbb R^m)$ is dense in $C_0^\infty(\mathbb R^n \times \mathbb R^m)$.
My attempt
It's clear that Proposition 1 holds for functions in $S$. Then, given $\varphi \in C_0^\infty(\mathbb R^n \times \mathbb R^m)$, there exists $\{\varphi_k\} \subset S$ such that $\varphi_k \xrightarrow{k\to\infty} \varphi$ in $C_0^\infty(\mathbb R^n \times \mathbb R^m)$. We have
$$ \langle T_x, \langle S_y \varphi_k(x,y) \rangle \rangle= \langle S_y \langle T_x, \varphi_k(x,y) \rangle \rangle \quad \forall k,$$
Since $\varphi_k(x,\cdot) \xrightarrow{k\to\infty} \varphi_k(x,\cdot)$ in $C_0^\infty(\mathbb R^m)$ for every $x \in \mathbb R^n$, we get by the continuity of the distributions, \begin{align}\label{some} \langle S_y, \varphi_k(x,y) \rangle \xrightarrow{k\to\infty}\langle S_y ,\varphi(x,y)\rangle, \quad \quad (1) \end{align} pointwise for every $x \in \mathbb R^n$.
I already know that $x \mapsto \langle S_y, \varphi_k(x,y) \rangle $ and $x \mapsto \langle S_y, \varphi(x,y) \rangle$ belong to $C_0^\infty(\mathbb R^n)$ , and, $\partial^\alpha \langle S_y, \varphi_k(x,y) \rangle= \langle S_y, \partial^\alpha_x \varphi_k(x,y)\rangle $ and $\partial^\alpha \langle S_y, \varphi(x,y) \rangle= \langle S_y, \partial^\alpha_x \varphi(x,y)\rangle$.
I want to show that \begin{align} \langle S_y, \varphi_k(\cdot,y) \rangle \xrightarrow{k\to\infty}\langle S_y ,\varphi(\cdot,y)\rangle, \end{align} in $C_0^\infty(\mathbb R^n)$ to conclude, but I'm not sure how to.
Any help of idea would be appreciated.
EDIT
$\{\varphi_n\}\subset C_0^\infty(\mathbb R^p)$ converges to $\varphi \in C_0^\infty(\mathbb R^p)$ if $\partial^\alpha \varphi_k$ converges to $\partial^\alpha \varphi$ uniformly for every multiindex $\alpha$.