This claim would be equivalent to saying that all linear functionals on the topological vector space $C_c^{\infty}(\mathbb{R}^n)$ are continuous.
This is not true, indeed, on any infinite dimensional topological vector space $X$ over a field $K$ there is a linear functional which is discontinuous.
Indeed, let $X$ be an infinite dimensional topological vector space, and let $\left\{e_n\right\}$ be a vanishing $(e_n\to 0)$ infinite sequence of linearly independent vectors on $X$, which may be easily constructed using continuity of the scalar product map. Define a linear functional $f$ on $X$ as any linear extension of the map
$$f(e_n):=n $$
Then the sequence $\left\{e_n\right\}\subset X$ is such that $e_n\to 0$, but $f(e_n)\not\to 0$.
For instance, in $C_c^{\infty}(\mathbb{R})$ (but it is the same in higher dimensions), recall that we say that a sequence of test functions $\left\{\varphi_n\right\}$ converges to $0$ iff their supports $\operatorname{supp}\varphi_n$ are all contained on some compact set $K\subset \mathbb{R}$ and $\varphi_n^{(k)}\to 0$ uniformly on $K$ for each order of differentiation $k\in \mathbb{N}$.
EDIT: The above is the usual topology that is set in the space of test functions $C_c^{\infty}(\mathbb{R}^n)$. In order to discuss whether
a functional $f:C_c^{\infty}(\mathbb{R}^n)\to \mathbb{R}$ is continuous
we first need to fix a topology on $C_c^{\infty}(\mathbb{R}^n)$, so I
assume you are referring to this one.
We may start from a function $0\neq \varphi \in C_c^{\infty}(\mathbb{R})$ and then set
$$e_n(x):=\frac{1}{n}\varphi\left(x+\frac{1}{n}\right) $$
so that the $\left\{e_n\right\}$ are linearly independent (since their supports are disjoint), $\operatorname{supp}e_n\subset \operatorname{supp}\varphi+[-1,0)$ and $e_n^{(k)}=\frac{1}{n}\varphi^{(k)}\left(x+\frac{1}{n}\right)\to 0$ uniformly for all $k$ as $n\to +\infty$. Thus, $\left\{e_n\right\}$ is a vanishing sequence of linearly independent vectors, and the functional $f$ defined above in the general case is a discontinuous linear functional on $C_c^{\infty}(\mathbb{R})$.
On the other hand, this is basically a sequence of bump functions which quickly reduces in overall size as $n\to +\infty$ and slowly shifts towards the left. Can you imagine a reasonable linear functional which blows up on such a sequence of functions?
As $n\to +\infty$, nothing meaningful 'blows up' here. I think this goes to show that you can usually expect your distributions to be continuous.
Take $\varphi_1 \in \mathcal{D}'(\mathbb{R}^{n-1})$ and $\varphi_2 \in \mathcal{D}'(\mathbb{R}).$
If $\varphi_2(0)=0$ then there exists $\psi \in \mathcal{D}'(\mathbb{R})$ such that $\varphi_2(x) = x \, \psi(x)$ giving
$$
\langle u, \varphi_1 \otimes \varphi_2 \rangle
= \langle u, \varphi_1 \otimes x \psi \rangle
= \langle x_n u, \varphi_1 \otimes \psi \rangle
= \langle 0, \varphi_1 \otimes \psi \rangle
= 0.
$$
Otherwise, take $\rho \in \mathcal{D}'(\mathbb{R})$ such that $\rho(0)=1,$ and set $\hat{\varphi}_2 = \varphi_2 - \varphi_2(0) \rho.$ Then $\hat{\varphi}_2(0) = 0$ so $\langle u, \varphi_1 \otimes \hat{\varphi}_2 \rangle = 0.$ Thus
$$
\langle u, \varphi_1 \otimes \varphi_2 \rangle
= \langle u, \varphi_1 \otimes (\hat{\varphi}_2 + \varphi_2(0) \rho) \rangle
= \langle u, \varphi_1 \otimes \hat{\varphi}_2 \rangle + \langle u, \varphi_1 \otimes \varphi_2(0) \, \rho \rangle \\
= \langle u, \varphi_1 \otimes \rho \rangle \varphi_2(0)
.
$$
Now, define $v_\rho \in \mathcal{D}'(\mathbb{R}^{n-1})$ by
$\langle v_\rho, \varphi_1 \rangle
= \langle\langle u(x',x_n), \rho(x_n)\rangle, \varphi_1(x')\rangle
.$
Then
$$
\langle u, \varphi_1 \otimes \rho \rangle \varphi_2(0)
= \langle\langle u(x',x_n), \rho(x_n)\rangle, \varphi_1(x')\rangle \, \langle \delta, \varphi_2 \rangle
= \langle v_\rho, \varphi_1\rangle \langle \delta, \varphi_2 \rangle
= \langle v_\rho \otimes \delta, \varphi_1 \otimes \varphi_2 \rangle.
$$
Thus,
$
\langle u, \varphi_1 \otimes \varphi_2 \rangle
= \langle v_\rho \otimes \delta, \varphi_1 \otimes \varphi_2 \rangle
$
and by uniqueness, we must have $u = v_\rho \otimes \delta.$
Best Answer
To prove continuity of $\phi$, it's enough to note that if $\phi_n\rightarrow\phi$ in $C_c^{\infty}(\mathbb{R}^2)$ then $$\psi_n=\phi_n(\cdot,0)\rightarrow\phi(\cdot,0)=\psi$$ in $C_c^{\infty}(\mathbb{R})$, which should follow from definitions (the derivatives of $\psi_n$ are partial derivatives of $\phi_n$).
The tensor product $T_1\otimes T_2$ of two distributions $T_1$ and $T_2$ is defined the same way as: $$\langle T_1\otimes T_2,\phi\rangle=\langle T_1, \langle T_2, \phi(x_1,\cdot)\rangle\rangle=\langle T_2, \langle T_1, \phi(\cdot,x_2)\rangle\rangle$$ It is the unique distribution which satisfies $$\langle T_1\otimes T_2,\phi_1\otimes\phi_2\rangle=\langle T_1,\phi_1\rangle\cdot\langle T_2,\phi_2\rangle$$ where $\phi_1\otimes\phi_2$ is the usual tensor product of functions: $$\big(\phi_1\otimes\phi_2\big)(x_1,x_2)=\phi_1(x_1)\phi_2(x_2)$$