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There are several ways one could interpret the word "constructivism" here, and the answer depends on what you meant by it.

Bishop-style constructivism is a generalization of Brouwerian intuitionistim, Russian constructivism, and classical mathematics. It is mathematics done without excluded middle (of course, you can still use excluded middle on those instances that you can prove to hold using other means) and general axiom of choice, but you still have countable choice. Thus, anything you prove in this setting is true in classical mathematics as well.

There are other forms of constructivism which are Bishop-style constructivism extended with additional principles and axioms. These additional principles often contradict classical logic, and so you get consequences that are classically false. Here are some examples:

In the internal language of the effective topos (an older name for this is Russian constructivism) the following are valid statements:

1. There are countably many countable subset of $\mathbb{N}$.
2. There is an increasing sequence in $[0,1]$ that has no accummulation point.
3. The Cantor space $2^\mathbb{N}$ and the Baire space $\mathbb{N}^\mathbb{N}$ are homeomorphic.
4. Every map $f : [0,1] \to \mathbb{R}$ is continuous.
5. There exists a continuous unboudned map $f : [0,1] \to \mathbb{R}$.
6. There is a covering of $\mathbb{R}$ by intervals $(a_n, b_n)_n$ with rational endpoints such that $\sum_{k = 1}^n |b_n - a_n| < 1$ for all $n \in \mathbb{N}$.
7. There is a subset $S \subseteq \mathbb{N}$ which is not finite and not infinite.
8. There exists an infinite binary rooted tree in which every path is finite.

In the internal language of the realizability topos $\mathsf{RT}(K_2)$ (an older name for this is Brouwerian intuitionisism) the following statements are valid:

1. Every map $f : X \to Y$ between complete separable metric spaces is continuous.
2. Every map $f : [0,1] \to \mathbb{R}$ is uniformly continuous.
3. Every map $f : \mathbb{R} \to \{0,1\}$ is constant, or equivalently, if $\mathbb{R} = A \cup B$ and $A \cap B = \emptyset$ then $A = \mathbb{R}$ or $B = \mathbb{R}$.

There are many other examples. I recommend taking the effort to get used to these amazing new worlds of mathematics.