Is a Cauchy sequence bounded in a metric space if the space is not complete?

I don't think a metric space needs to be complete for a Cauchy sequence to be bounded. Being complete means the sequence converges to a point in the space, but the standard proof for Cauchy sequence being bounded does not require you to know what the sequence converges to. For that reason, I think it doesn't matter whether the metric space is complete or not.

A more conceptual way to see it: Let $$X$$ be a metric space, $$x_n$$ a Cauchy sequence in $$X$$, and let $$\overline X$$ be its completion. We can isometrically embed $$X$$ into $$\overline X$$, and the image $$\overline x_n$$ of the sequence $$x_n$$ will still be a Cauchy sequence, since its elements still have the same distance from each other. But this sequence converges, since we're in a complete space, so it must be bounded. And since the distances between the $$x_n$$ are the same as the ones between the $$\overline x_n$$, $$x_n$$ must also be bounded.

Yes it is.

Let $$(x_n)_{n \in \mathbb{N}}$$ be a Cauchy sequence in a metric space $$(E,d)$$. By definition, there exists $$N \in \mathbb{N}$$ such that for every $$p \geq N$$, you have $$d(x_p,x_N)\leq 1$$

Let $$M = \max \lbrace d(x_i,x_N), 0 \leq i \leq N-1 \rbrace + 1$$

You have then that for every $$i \in \mathbb{N}$$, $$d(x_i,x_N) \leq M$$

so for every $$i \in \mathbb{N}$$, $$x_i \in B(x_N, M)$$

(where $$B(x_N, M)$$ denotes the ball centered at $$x_N$$ of radius $$M$$) ; i.e. the sequence is bounded.

If $$(x_n)_{n\in\Bbb N}$$ is a Cauchy sequence in a metric space $$(X,d)$$, then there is some $$N\in\Bbb N$$ such that $$n\geqslant N\implies d(x_m,x_n)<1$$. Therefore$$\{x_n\mid n\geqslant N\}\subset B_1(x_N).$$Now, let $$M=\max\{d(x_N,x_k)\mid k; it exists, since it is the maximum of a finite set. Then$$\{x_n\mid nand therefore, since $$M+1>1$$,$$\{x_n\mid n\in\Bbb N\}\subset B_{M+1}(x_N).$$

Hint:For fixed $$y$$ the sequence $${d(x_n,y)$$ is a Cauchy sequence of real numbers. Hence it is bounded.