I think I've come up with a proof.

Assume that $u \in \dot{B}^{s}_{p,r}(\mathbb R^d)$ with $s < d/p$ and $p,r \in [1,+\infty]$. Write $\dot{\Delta}_j = \varphi(2^{-j}D)$, $j \in \mathbb Z$.

**Approximate identity convergence**

Let $(k_{\lambda})_{\lambda > 0} \subset S(\mathbb R^d)$ be any approximate identity that lies in Schwartz space, i.e.

- $\int_{\mathbb R^d} k_{\lambda} = 1$
- $\sup_{\lambda > 0} ||k_{\lambda}||_{L^1} < +\infty$
- $\forall \delta > 0$ : $\lim \limits_{\lambda \to \infty} \int_{|x| \geq \delta}||k_{\lambda}||_{L^1} = 0$

Trivially,

$$||\chi(2^nD)(u * k_{\lambda})||_{L^{\infty}} = ||(\chi(2^nD)u) * k_{\lambda}||_{L^{\infty}} \leq ||\chi(2^nD)u||_{L^{\infty}} \cdot ||k_{\lambda}||_{L^1}$$

Using Young-Convolution inequality,
$$2^{js}||\dot{\Delta}_j(u * k_{\lambda})||_{L^p} = 2^{js}||(\dot{\Delta}_ju) * k_{\lambda}||_{L^p} \leq 2^{js}||\dot{\Delta}_ju||_{L^p} \cdot ||k_{\lambda}||_{L^1}$$

Hence, $u * k_{\lambda} \in \dot{B}^{s}_{p,r}(\mathbb R^d)$ and it is uniformly bounded in $\lambda$. By the Fatou property, the limit (along some subsequence) as $\lambda \to +\infty$ exists in $\dot{B}^{s}_{p,r}(\mathbb R^d)$.

Since $\dot{B}^{s}_{p,r}(\mathbb R^d)$ is embedded into $S'(\mathbb R^d)$ (which is Hausdorff) and $u * k_{\lambda}$ converges to $u * \delta = u$ in $S'(\mathbb R^d)$, the limit in $\dot{B}^{s}_{p,r}(\mathbb R^d)$ must be $u$.

**Smooth cut-off convergence**

If $f$ is Schwartz, then
$$||u \cdot f||_{\dot{B}^{s}_{p,r}} \leq C [ ||f||_{L^{\infty}} + ||f||_{\dot{B}^{d/p}_{p,\infty}} + ||f||_{\dot{B}^{0}_{\infty,\infty}} ] \cdot ||u||_{\dot{B}^{s}_{p,r}}$$
thanks to paraproduct continuity and the embedding $\dot{B}^{s}_{p,r} \subset \dot{B}^{s-d/p}_{\infty,r}$.

Moreover,
$$||f||_{\dot{B}^{0}_{\infty,\infty}} \leq C ||f||_{L^{\infty}}$$

Now, we need to study
$$||f||_{\dot{B}^{d/p}_{p,\infty}} = \sup_{j \in \mathbb Z} (2^j)^{d/p} ||\dot{\Delta}_j f||_{L^p}$$
We must use something which looks like Bernstein inequality, but with a fractional derivative $\partial^{d/p}$ (defined by means of the Fourier transform).
I think we can prove
$$(2^j)^{d/p} ||\dot{\Delta}_j f||_{L^p} \leq C ||\partial^{d/p} \dot{\Delta}_j f||_{L^p} \leq C ||\partial^{d/p} f||_{L^p}$$

If $f$ is a smooth cut-off, $f(x) = \chi(x/R)$ with $0 \leq \chi \leq 1$, $\chi \in C^{\infty}_c(\mathbb R^d)$, then
$$||u \cdot \chi(x/R)||_{\dot{B}^{s}_{p,r}} \leq C [ ||\chi||_{L^{\infty}} + ||\partial^{d/p} \chi||_{L^p} ] \cdot ||u||_{\dot{B}^{s}_{p,r}}$$
so it is uniformly bounded for $R > 0$. Thanks to Fatou property, the limit (along some subsequence) as $R \to +\infty$ exists in $\dot{B}^{s}_{p,r}(\mathbb R^d)$.

Finally, the limit, denoted by $v$, must satisfy
$$\langle v, \varphi \rangle = \langle u, \varphi \rangle$$
for all test functions $\varphi \in C^{\infty}_c(\mathbb R^d)$. Hence, $u$ and $v$ must differ by a constant which is exactly zero.

For references on theorems used here, look at Bahouri "Fourier Analysis and Nonlinear Partial Differential Equations"