Proof Verification: Differentiability implies continuity. Proof:
Let $x$ be a real number, since $f(x)$ is differentiable, or $\lim_{x \to 
 a} \frac{f(x)-f(a)}{x-a}$ exists and is equal to $f'(a)$. So, for any $\epsilon' > 0$ there exists a $\delta > 0$, such that $0<|x-a| < \delta$ 
$\implies |\frac{f(x)-f(a)}{x-a} - f'(a)| < \epsilon'$
Using the Triangle inequality, we get
$ |\frac{f(x)-f(a)}{x-a}| < \epsilon' + |f'(a)|$
$\implies |f(x)-f(a)| < \epsilon' \delta + |f'(a)| \delta$
Here's where I start getting unsure whether my proof is correct or not. This inequality is true for $\epsilon'$ = $\frac{\epsilon}{\delta'} - |f'(a)|$ where $\delta' < \delta$ such that $\frac{\epsilon}{\delta'} >  |f'(a)|$
This gives us 
$ |f(x)-f(a)| < \epsilon$
The logic behind the proof is that if someone gives asks me to find an appropriate delta for $\epsilon = h$, I'll find a delta for $\epsilon' = h$, say that delta is equal to $n$. Next I solve for $\epsilon'$ using $\epsilon' = \frac{h}{n} - |f'(a)|$. If on solving I get a negative $\epsilon'$, I decrease $\delta'$ to a value $n_0$ such that I get a positive value for $\epsilon'$ and then get the corresponding delta, $n_1$. The minimum of $n_1$ and $n_0$ would be the required delta.
Please let me know where exactly the proof starts to go wrong. While alternate proofs are appreciated, the main goal here is understanding why this proof is wrong.
EDIT: After thinking about the problem for a while and reading a few of the answers, I found a way to communicate the idea more effectively. 
Continuing from $|f(x)-f(a)| < (\epsilon' + |f'(a)|) \delta$
Now we need to show that we can represent any positive real number, $\epsilon$, by taking appropriate values of $\delta$ and $\epsilon'$. Fix $\epsilon' = 1$. Now $\delta$ is just $\frac{\epsilon}{1+|f'(a)|}$. To complete our answer, let one value of $\delta$ for $\epsilon' = 1$ be $\delta_0$. Our final answer would be $\delta = \min(\delta_0,\frac{\epsilon}{1+|f'(a)|})$
 A: Or just see that
$$\lim_{h\rightarrow 0}f(x+h)-f(x)=\lim_{h\rightarrow 0}\underbrace{\frac{f(x+h)-f(x)}{h}}_{\rightarrow f'(x)}h=0.$$
A: Not quite right. One of the issues is:

I define $\epsilon'$ as $\frac \epsilon \delta $

As it stands, this doesn't make sense, because you started the proof by taking an arbitrary $\epsilon'$. Perhaps you mean something like: Since this inequality is true for any $\epsilon '$, it is true in particular for  $\frac \epsilon \delta$ ...
But this doesn't quite make sense either; What's the $\delta$ on the right-hand side? This $\delta $ may depend on the $\epsilon '$ you initially choose. 

Try this alternative:
For any $\epsilon> 0$ there is a $\delta' > 0$, such that whenever $0 < |x-a| < \delta'$, one has $\left|\frac{f(x)-f(a)}{x-a}\right| < |f'(a)| + \epsilon$. In particular, there is a $\delta_0>0$ such that $\left|\frac{f(x)-f(a)}{x-a}\right| < |f'(a)| + 1$
Now, choose $\delta = \min \left(\delta_0, \frac 1 {|f'(a)+1|}\right) $
Can you complete the proof from here?
A: The heart of your argument is basically the following. Give me some $\epsilon$, and I have to find some $\delta$ so that $|f(x)-f(a)|<\epsilon$ when $|x-a|<\delta$. To start with, I'll just take any $\delta$, it doesn't matter. Now, within the range $[a-\delta, a+\delta]$, the function $x\to\frac{f(x)-f(a)}{x-a}$ is bounded, because it's convergent at $a$ and convergent functions are locally bounded - this is a theorem which you essentially spend the first couple of lines of your proof re-proving by way of the value $\epsilon'$. So say $\lvert\frac{f(x)-f(a)}{x-a}\rvert<M$ on the interval $[a-\delta, a+\delta]$. Well then certainly $|f(x)-f(a)|<M\delta$, and by choosing a sufficiently small $\delta$ we can make that less than $\epsilon$ (since the $M$ only gets smaller as $\delta$ gets smaller).
If you notice, in your own proof, the $\epsilon'$ is a bit pointless. Since at the end you're just going to say "and now take $\delta$ as small as is necessary to make this true", you may as well initially take $\epsilon'=10^{100}$. All that matters is that $\epsilon'$ is finite and that $\delta$ can be made arbitrarily small, in other words, that the difference quotient is locally bounded. The exact value of the derivative at $a$ also doesn't matter.
A: We can simply use the equivalent definition of differentiability
$$f(a+h)=f(a)+f'(a)\cdot h +o(h) \implies  \lim_{x\to a} f(x)=\lim_{h\to 0} f(a+h)=f(a)$$
A: $ |\frac{f(x)-f(a)}{x-a}| < \epsilon' + |f'(a)|$ 
so
$ |f(x)-f(a)| <( \epsilon' + |f'(a)|)|x-a|$
then for $x\to a$ you have that 
$|f(x)-f(a)|\leq 0$
A: I try to put your arguments in the right order.
Proof:
Assume that $f$ is differentiable in $a$. So $f'(a)$ exists.
To prove that $f$ is continous in $a$ choose an arbitrary  $\varepsilon>0.$ Now you can choose $\delta_1$ such
$$0<\delta_1<\frac{\varepsilon}{|f'(a)|}, \text{ if } f'(a)\ne 0$$
$$\delta_1=1, \text{ if } |f'(a)|=0.$$
So we have
$$0<|f'(a)|<\frac{\varepsilon}{\delta_1}$$
and define
$$\varepsilon_2=\frac{\varepsilon}{\delta_1}-|f'(a)|>0$$
Because f is differentiable in $a$ we can find a $\delta_2$ such that
$$|\frac{f(x)-f(a)}{x-a} - f'(a)| < \varepsilon_2,\; \forall x: 0<|x-a|<\delta_2$$
Using the triangle inequality, we get
$$ |\frac{f(x)-f(a)}{x-a}| < \varepsilon_2 + |f'(a)|,\; \forall x: 0<|x-a|<\min(\delta_1,\delta_2)$$
So we set 
$$\delta=\min(\delta_1,\delta_2)$$
and get
$$| f(x)-f(a)| < (\varepsilon_2 + |f'(a)|) \delta_2<\frac{\varepsilon}{\delta_1}\min(\delta_1,\delta_2)\le \varepsilon,\; \forall x: 0<|x-a|<\delta$$
