Does Cauchy continuity imply uniform continuity? [No.] It is well known that if $X$ is a first countable topological space and $Y$ is a topological space, then $f : X \rightarrow Y$ is continuous iff
$$\forall x \in {\rm map}(\mathbb{N},X),\forall p \in X \quad x_{n} \rightarrow p \Rightarrow f(x_{n}) \rightarrow f(p)$$
It is also well known that if $X$ and $Y$ are metric spaces and $f : X \rightarrow Y$ is uniformly continuous, then $f$ maps Cauchy sequences to Cauchy sequences. 
By analogy it seems plausible that if a function between metric spaces maps Cauchy sequences to Cauchy sequences then it must be uniformly continuous. However mimicking the proof of the analogous result for continuous maps doesn't work, which makes me think the result if false. Does anyone know any counterexamples?
Also on the uniform continuity wikipedia page, it says that the result is true if $X$ and $Y$ are subsets of $\mathbb{R}^{n}$. EDIT: It actually doesn't say this, I misread the page.
 A: Take for $X$ the disjoint union of a sequence $A_n$ of two point sets. Set $d(p,q)=1/n$ if $p$ and $q$ are different points in $A_n$ and $d(p,q)=1$ if $p\in A_n$ and $q\in A_m$ with $m\not=n$.  Every Cauchy sequence in $X$ is eventually constant.  Consider the identity mapping from $X$ with this metric to $X$ with the discrete metric.
A: Several people have already given examples to the effect that preservation of Cauchyness is not enough to prove that a map is uniformly continuous. It is still possible however, to characterize uniform continuity in terms of sequences (for metric spaces only, for uniform spaces you would need nets (or filters)). In case you are interested here goes the result.
Theorem: Let $f:X\to Y$ be a map between metric spaces (both metrics denoted by $d$). Then $f$ is uniformly continuous iff, for every pair of sequences $(x_n)$ and $(z_n)$ in $X$ such that $d(x_n, z_n)$ converges to $0$, then $d(f(x_n), f(z_n))$ converges to zero.
Proof: exercise left to the reader.
A: Well, here is [most probably another] Wikipedia page: http://en.wikipedia.org/wiki/Cauchy-continuous_function. HTH.
Alternatively, you may use the function $f:\mathbb{R\rightarrow\mathbb{R}}$ , expressed
by $f\left(x\right)=x^{2}$ to show that it is possible for a continuous
function to send Cauchy sequences to Cauchy sequences without being
uniformly continuous.
As for the continuity question: the function must be continuous. For,
embed $Y$ into its completion, say $Y^{\sim}$, let $p\in X$ , and
let $\left(x_{n}\right)$ be a sequence in $X$ converging to $p$
. Then the sequence $\left(x_{1}, p, x_{2}, p,...\right)$
is Cauchy in $X$ , isn't it ? And, therefore, its image by $f$ should
be a Cauchy sequence in $Y$ , hence convergent in $Y^{\sim}$ .
Yet, that image contains a constant subsequence... isn't it ?
A: No it's not true.
f(x) = x^2 on whole real line.
It maps Cauchy sequences to Cauchy sequences but it's not uniformly continuous on the whole real line.
