Can someone check my proof if a function $f:{R\mapsto R}$ is differentiable at all points, it is continuous throughout as well?

Proof:

Let $x$ be a real number, since $f(x)$ is differentiable, or $\lim\_{h \to 
 0} \frac{f(x+h)-f(x)}{h}$ exists and is equal to $f'(x)$. So, for any $\epsilon > 0$ there exists a $\delta > 0$, such that $0<|h - 0| < \delta$ 

$\implies |\frac{f(x+h)-f(x)}{h} - f'(x)| < \epsilon$

$\implies |\frac{f(x+h)-f(x)-f'(x).h}{h}| < \epsilon$

$\implies |f(x+h)-f(x)-f'(x).h| < |h|\epsilon$

$\implies -|h|\epsilon<f(x+h)-f(x)-f'(x).h < |h|\epsilon$

$\implies f'(x).h-|h|\epsilon<f(x+h)-f(x) < f'(x).h+|h|\epsilon$

Let $\epsilon'$ be min($f'(x).h+|h|\epsilon - f(x+h)+f(x), f'(x).h-|h|\epsilon-f(x+h)+f(x) $), then we can say,

$\implies -\epsilon'<f(x+h)-f(x) < \epsilon'$

$\implies |f(x+h)-f(x)| < \epsilon'$

$\implies \lim\_{h \to 0}f(x+h)$ exists and is equal to $f(x)$

$\implies f$ is continuous at $x$

NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.