I read the following:
Given $E$ and $F$ as metric spaces and $\hat{E}$ and $\hat{F}$ as their completion, any uniformly continuous function $\phi : E \rightarrow F$ admits a unique uniformly continuous extension $\hat{\phi}: \hat{E} \rightarrow \hat{F}$. Furthermore, if $\hat{E}$ is compact and $\phi$ is surjective, then $\hat{\phi}$ is surjective.
I would like to ask, is there a proof for this statement that uses the contrapositive ? i.e.
$\hat{\phi}$ is not surjective $\implies$ $\hat{E}$ is not compact or ${\phi}$ is not surjective
To show $[\hat{\phi}$ is not surjective $\implies$ $\phi$ is not surjective $]$ is one part, but how about the other part of the statement, i.e. that if $\hat{\phi}$ is not surjective, it can be that $\phi$ is surjective but $\hat{E}$ is not compact ?
Your contrapositive has the form $p\to(q\lor r)$. The usual way to prove such an implication is to prove $(p\land\neg q)\to r$ or $(p\land\neg r)\to q$. In this case that means proving either that if $\widehat\varphi$ is not surjective and $\varphi$ is, then $\widehat E$ is not compact, or that if $\widehat\varphi$ is not surjective and $\widehat E$ is compact, then $\varphi$ is not surjective. The first of these possibilities looks a bit more promising, so let’s start there.
If $\widehat\varphi$ is not surjective, fix $y\in\widehat F\setminus\widehat\varphi[\widehat E]$, and let $\langle y_n:n\in\Bbb N\rangle$ be a sequence in $F$ converging to $y$. If $\varphi$ is surjective, there is a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ in $E$ such that $\varphi(x_n)=y_n$ for each $n\in\Bbb N$. If some subsequence of $\sigma$ converged to some $p\in\widehat E$, we would have $\widehat\varphi(p)=y$, which is impossible, so $\sigma$ has no convergent subsequence in $\widehat E$, and therefore $\widehat E$ is not compact.
What about the other possibility? As before, if $\widehat\varphi$ is not surjective, fix $y\in\widehat F\setminus\widehat\varphi[\widehat E]$, and let $\langle y_n:n\in\Bbb N\rangle$ be a sequence in $F$ converging to $y$. If $\widehat E$ is compact, then $\widehat\varphi[\widehat E]$ is compact and hence closed in $\widehat F$, so $\widehat F\setminus\widehat\varphi[\widehat E]$ is an open nbhd of $y$, and there is an $m\in\Bbb N$ such that $y_n\in\widehat F\setminus\widehat\varphi[\widehat E]$ whenever $n\ge m$. And each $y_n\in F$, so in fact $y_n\in F\setminus\widehat\varphi[\widehat E]\subseteq F\setminus\varphi[E]$ whenever $n\ge m$, and $\varphi$ is therefore not surjective.
So in this case both of the natural alternatives for proving the contrapositive in question turn out to be pretty straightforward.