Q:

Why when the implication is false, its converse is also false in this example?

As we all know: 

To prove that an implication is true,just prove that its converse is false.

For example,from elementary calculus we know that the assertion
"If a function is continuous, then it is differentiable"
is false. This allows us to reach at the correct conclusion that its converse,
"If a function is differentiable, then it is continuous"
is true, as indeed it is. 
But wait a minute! The implication"
If a function is differentiable, then it is not continuous"
is completely false. So we could conclude that its converse 
"If a function is not continuous, then it is differentiable"
should be true, but in fact the converse is also completely false!
So something has gone wrong here. But where?

A:

It is logically correct to say that 
$$ (p \implies q)\quad OR \quad (q \implies p)$$
It is also logically correct to say
$$ [(p(x)\implies q(x))\quad OR \quad (q(x) \implies p(x))] \quad \forall x \in X$$
However, it is not logically correct to say that 
$$ [(p(x)\implies q(x)) \quad \forall x \in X] \quad OR \quad [(q(x)\implies p(x)) \quad \forall x \in X] $$
Your statement about differentiable functions has the latter form (and is not correct):  
$$[(\mbox{$f$ differentiable} \implies \mbox{$f$ not continuous})\forall f \in F]\quad OR \quad [(\mbox{$f$ not continuous} \implies \mbox{$f$ differentiable})\forall f \in F]$$
where $F$ is the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.  On the other hand, if we take any specific function $f \in F$, such as $f(t) =e^t$ for all $t\in \mathbb{R}$, then we indeed get a correct statement: 
$$(\mbox{$e^t$ differentiable} \implies \mbox{$e^t$ not continuous})\quad OR \quad (\mbox{$e^t$ not continuous} \implies \mbox{$e^t$ differentiable})$$ and indeed the latter is true.

Aside: Proof of the more general logical statement: For any three propositions $p,q,r$ we have
$$ (p\implies q) \quad OR \quad (q \implies r)$$
Proof: 
\begin{align}
(p\implies q) \quad OR (q \implies r) &= (\lnot p \quad OR \quad q) \quad OR \quad (\lnot q \quad OR \quad r)\\
&= (\lnot p) \quad OR\quad  (q) \quad OR \quad (\lnot q) \quad  OR\quad  r\\
&= True
\end{align}

