Let $h : U \times \mathbb{R}^n \to \mathbb{R}^m$ be a continuously differentiable function, where $U$ is an open of $\mathbb{R}^p$, to be understood as the space of parameters. Assume that for all $(u,x) \in U \times \mathbb{R}^n$ such that $h(u,x) = 0$, the partial differential with respect to the second variable, $$ y \mapsto \mathrm{d}h(u,x)(0,y), $$ is surjective, namely all roots of $h$ are regular. Then, by the preimage theorem (or regular level set theorem), we know that $L \triangleq h^{-1}(\{0\})$ is a differentiable submanifold of $U \times \mathbb{R}^n$, and for all $u \in U$, $$ M(u) \triangleq h(u,\cdot)^{-1}(\{0\}) = \{x \in \mathbb{R}^n, ~h(u,x)=0\}, $$ is differentiable submanifold of $L$, the "$u$-slice of $L$" in some sense.
Considering $M$ as a set-valued function (correspondence) and assuming its values are non-empty, is $M$ continuous (upper and lower hemicontinuous)?
Lower hemicontinuity
Let $u_0 \in U$ and $V$ be an open of $\mathbb{R}^n$ intersecting $h^{-1}_{u_0}(\{0\})$. Let then $x_0 \in \mathbb{R}^n$ such that $h(u_0,x_0) = 0$ and $(u_0, x_0) \in V$. We will prove that there exists $W$ a neighborhood of $u_0$ such that $h^{-1}_u \cap V \neq \emptyset$ for all $u \in W$.
For clarity, we define the following surjective (by hypothesis) map, $$ \lambda : x \in \mathbb{R}^n \mapsto \mathrm{d}(u_0, x_0)(0, x). $$ We note $S$ a supplement of $\ker \lambda$, namely a subspace of $\mathbb{R}^n$ such that $S \oplus \ker \lambda = \mathbb{R}^n$. We let $(f_1, \dots, f_p)$ be a basis of $S$ and $(f_{p+1}, \dots, f_n)$ be a base of $\ker \lambda$, while $(e_1, \dots, e_n)$ is the canonical basis of $\mathbb{R}^n$. Consider now the change of basis map, $$ \Phi : e_i \mapsto f_i, $$ and define for $s \in \mathbb{R}^m, k \in \mathbb{R}^{n-m}, u \in U$, $$ \tilde{h}(u,s,k) = h(u,\Phi(s,k)). $$
Defining $(s_0, k_0) = \Phi^{-1}(x_0)$, we have for all $(s,k) \in \mathbb{R}^n, u \in \mathbb{R}^p$, $$ \begin{align*} \mathrm{d}\tilde{h}(u_0,s_0,k_0)(u,s,k) &= \mathrm{d}h(u_0,x_0)(u, \mathrm{d}\Phi(s_0, k_0)(s,k)) \\ &= \mathrm{d}h(u_0,x_0)(u, \Phi(s,k)), \end{align*} $$ so that, $$ \tilde{\lambda} : s \in \mathbb{R}^p \mapsto \mathrm{d}\tilde{h}(u_0,s_0,k_0)(0,s,0) = \mathrm{d}h(u_0,x_0)(0,\Phi(s,0)), $$ is invertible. Applying the implicit function theorem, there exists an open neighborhood $\Omega$ of $(u_0,x_0)$, a neighborhood $W$ of $(u_0, k_0)$ and a continuously differentiable function $\phi$ defined on $W$ to $\mathbb{R}^p$, such that, $$ ((u,s,k) \in \Omega \text{ and } \tilde{h}(u,s,k)) \quad \iff \quad ((u,k) \in W \text{ and } s = \phi(u,k)). $$ Without loss of generality, $W$ is a ball centered on $(u_0, k_0)$. We call $\tilde{W} = W \cap (U \times \{0\})$, and see that for all $u \in \tilde{W}$ open neighborhood of $u_0$, $$ h(u,\Phi(\phi(u,0),0)) = \tilde{h}(u,\phi(u,0),0) = 0. $$ Without generality, we can assume that $\Phi(\phi(W,0),0) \subset V$ (by continuity of $\phi$ and $\Phi$), therefore $u \mapsto h^{-1}_u(\{0\})$ is lower semicontinuous.
Upper hemicontinuity
As it turns out we are lacking hypotheses here. Indeed, consider, $$ h : (u,x,y) \in \mathbb{R}^3 \mapsto y-ux^2, $$ which gives, $$ u \mapsto h^{-1}_u(\{0\}) = \{ (x,ux^2), x \in \mathbb{R} \}. $$ This function is not upper hemincontinuous: let $V = \mathbb{R} \times (-1,1)$ which contains $h^{-1}_0(\{0\})$, yet for all $u \neq 0$ the point $(\frac{1}{\sqrt{u}},1)$ belongs to $h^{-1}_u(\{0\})$ but not to $V$.
However, if we assume that $u \mapsto h^{-1}_u(\{0\})$ is compact-valued, then by the closed graph theorem, it is indeed upper hemicontinuous.