Uniform Continuity implies Continuity Let $f$ be a function from a metric space $X$ to a metric space $Y$. Show that if $f$ is uniformly continuous on $X$ then $f$ is continuous on $X$. Show that the converse is not true.
Uniform continuity definition:
$∀ε>0:∃δ>0:∀p,q∈X:d_{X}(p,q)<δ⟹d_{Y}(f(p),f(q))<ε$
Continuity definition:
$∀ε>0:∀p∈X:∃δ>0:∀q∈X:d_{X}(p,q)<δ⟹d_{Y}(f(p),f(q))<ε$
I wonder if these proofs are correct and if they are formal enough?

Uniform continuity ⟹ Continuity


*

*Let $f$ be uniformly continuous. Fix $ε_{0}$, obtain $δ_{0}(ε_{0})$ (as a function of $ε_{0}$), fix any $p_{0}$ and $q_{0}$ and we know that:


$d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$


*If we want to prove that $f$ is continuous at $p_{0}$, we fix $ε_{0}$ and we pick the same $δ_{0}$ as above and fix any $q_{0}$ and we are assured by 1. that


$d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$


*Since $p_{0}$ was arbitrary, $f$ is continuous at $X$.



Continuity $\color{red}{\implies}$ Uniform continuity


*

*Let $f$ be continuous. Fix $ε_{0}$, fix any obtain $p_{0}$ and $p_{1}$, obtain both $δ_{0}(ε_{0},p_{0})$ and $δ_{1}(ε_{0},p_{1})$ (not necessarily equal), fix any $q_{0}$ and $q_{1}$ and we know that:


$d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$
$d_{X}(p_{1},q_{1})<δ_{1}⟹d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}$


*For $f$ to be uniformly continuous, if we fix $ε_{0}$, obtain $δ$, fix any $p_{0}$, $p_{1}$, $q_{0}$ and $q_{0}$ we must have that:


$d_{X}(p_{0},q_{0})<δ⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}$
$d_{X}(p_{1},q_{1})<δ⟹d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}$


*But those two implications need not hold for the same $δ$ as shown above. Suppose we pick $δ_{0}$ and w.l.o.g. we suppose $δ_{0}>δ_{1}$. Then it could be the case that


$δ_{1}<d_{X}(p_{1},q_{1})<δ_{0}$
but we cannot conclude ⟹$d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}$.
Therefore, continuity does not imply uniform continuity.
 A: You cannot usually prove than an implication fails; what you do is provide a counterexample. Your argument for continuity does not imply uniform continuity cannot work because (as you showed before) there are functions which are both. 
For a continuous but not uniformly continuous function, intuitively  you need a function that has arbitrarily high slopes (i.e., unbounded derivative if it has one). An easy example is $f(x)=x^2$ with $X=Y=\mathbb R$. We need to show that this fails the definition of uniform continuity: that is, it satisfies its negation, which says
$$\tag{1}
\exists\varepsilon>0, \forall\delta>0,\ \exists p,q:\ d(p,q)<\delta\ \text{ and }d(f(p),f(q))>\varepsilon.
$$
For this function in particular we can fix any $\varepsilon$, say $\varepsilon=1$. Given $\delta>0$, we can take $p=k$, $q=k+\delta/2$ for some integer $k$ (to be determined). Then $d(p,q)=|p-q|=\delta/2$, and
$$
d(f(p),f(q))=d(p^2,q^2)=|(k+\delta/2)^2-k^2|=2k\delta+\delta^2/4>2k\delta.
$$
So if we choose $k>1/2\delta$, $(1)$ is satisfied. 
