Let $h : V \to \mathbb{R}$ be an analytic function where $V \subseteq \mathbb{R}$ is open. This means by definition that for all $c$ in the domain of $h$, there is some neighbourhood $U \subseteq V$ of $c$ such that $\forall x \in V (\sum\limits_{n = 0}^\infty \frac{h^{(n)}(c)}{n!} (x - c)^n = h(x))$. We assume that $V$ is connected and that there is some $c \in V$ such that for all $n$, $h^{(n)}(c) = 0$. Can we constructively show that $h = 0$?
We see that $U = \{c \in V \mid \forall n (h^{(n)}(c) = 0)\}$ and $W = \{c \in V \mid \exists n (h^{(n)}(c) \neq 0)\}$ are both open sets, where $\neq$ is the apartness relation if we're working constructively.
Classically, therefore, we see that $U$ and $W$ are complements. We see by connectedness that either $U = \emptyset$ or $W = \emptyset$. So since we have $c \in U$, we conclude that $W = \emptyset$ and therefore $h = 0$.
However, constructively, this does not work because in general, there is no reason to think we have $U \cup W = V$. It is certainly true that $\{c \in V \mid c \notin W\} = U$, but this is not strong enough to show that $U \cup W = V$.
For the constructive proof, we will limit ourselves a bit and require $V = (a, b)$ where $a < b$.
We can try again, this time using compactness. The basic notion here is a "single-series interval", which is an interval $U = (k - \epsilon, k + \epsilon) \subseteq V$ where for all $x \in U$, $h(x) = \sum\limits_{i = 0}^\infty \frac{h^{(n)}(k)}{k!} (x - k)^n$. It can be shown in a fairly straightforward and constructive way that a single-series interval has the property that if there is some $w \in U$ where for all $n$, $h^{(n)}(w) = 0$, then $h|_U = 0$. This can be done using a standard argument establishing the radius of convergence of the Taylor series centered at $w$ and then steadily walking the center for a Taylor expansion towards $k$. Furthermore, any interval subset of a single-series interval is also single-series, so a non-trivial intersection of 2 single-series intervals is single-series.
Then in particular, consider $\mathcal{F} = $ the set of all single-series intervals. We see that $\mathcal{F}$ is an open cover of $V$. This follows directly from the fact that $h$ is analytic on $V$.
Then consider the interval $[a + \epsilon, b - \epsilon]$ for some small $\epsilon > 0$ - small enough that $c \in [a + \epsilon, b - \epsilon]$. This interval is compact by the Heine-Borel theorem, and it is covered by $\mathcal{F}$. Therefore, we may take $U_1, \ldots, U_n \in \mathcal{F}$ such that $[a + \epsilon, a - \epsilon] \subseteq U_1 \cup \cdots \cup U_n$. I claim that $h$ is zero on each $U_i$.
To prove this, I do a rather strange induction. I show that for all $j \leq n$, we can come up with a list $i_1, \ldots, i_{j}$ of distinct indices such that $h$ is zero on each $U_{i_\ell}$, $1 \leq \ell \leq j$. The base case we shall take to be $j = 1$: in this case, pick some $i_1$ such that $c \in U_{i_1}$. For the inductive step, we have $[a + \epsilon, b - \epsilon] \subseteq (U_{i_1} \cup \cdots \cup U_{i_j}) \cup ($the union of the other $U$s). Therefore, since $(U_{i_1} \cup \cdots \cup U_{i_j})$ and the union of the other $U$s not yet selected both contain at least one point, by connectedness they overlap. Pick a $U_k$ which overlaps with $U_{i_\ell}$. Then $h$ is zero on $U_k$, so let $i_{j + 1} = k$.
Thus, we see that $h$ is zero on $[a + \epsilon, b - \epsilon]$ for all sufficiently small $\epsilon$, hence is zero on $(a, b) = \bigcup\limits_{\epsilon > 0} [a + \epsilon, b - \epsilon]$.
In the second proof, the only nonconstructive principle we apply is the Heine-Borel theorem (which is far weaker than excluded middle).
Can we find a purely constructive proof of this fact which doesn't rely on Heine-Borel?