An operator $C$ is bounded iff the set {$\|Cx\|:\|x\|\leq 1$} is bounded $\Leftrightarrow$ there is a $M<\infty:\|Cx\|\leq M\|x\|$ for every $x\in U$.
Let $ε>0$. If $x,y\in U:\|x-y\|<ε/M$, then $\|Cx-Cy\|\leq M\|x-y\|<ε$. Thus $C$ is not only continuous but uniformly continuous also.
So, a bounded operator is always continuous on norm-spaces. Banach space is a norm-space which is complete, thus things are not different there.
HINT: Imagine for a moment that an extension $T$ of $T_0$ exists and take $x\in E\setminus E_0$. Since $E_0$ is dense, you can approximate $x$ with a sequence $x_n\in E_0$. Since our "imaginary" operator $T$ is continuous, it must hold that $$\tag{1}Tx=\lim_{n\to \infty} T_0 x_n.$$
Now go back to reality, where $T$ does not exist yet. You need to construct it. The formula (1) gives you an obvious candidate, but you have to check that it makes sense at all points $x\in E$ and that it is independent of the choice of an approximating sequence $x_n$.
EDIT. Никита Васильев asked for more details in the comment section. Ok, here they are; we want (1) to be a consistent definition of an operator $T\colon E\to F$. For that, we need two things; first, the limit must exist, and second, if we choose another sequence $x_n'\in E_0$ such that $x_n'\to x$, then
$$\tag{2}
\lim_{n\to \infty} T_0x_n'= \lim_{n\to \infty} T_0 x_n.$$
To prove the first thing, we observe that the boundedness of $T_0$ gives
$$
\lVert T_0x_n-T_0 x_m\lVert \le C\lVert x_n-x_m\rVert.$$
From this it immediately follows that $(T_0 x_n)$ is Cauchy, because $x_n$ is. And since $F$ is complete, this assures that the limit exists. To prove the second thing, we note that
$$
\lVert T_0 x_n- T_0x_n'\lVert\le C\lVert x_n-x_n'\rVert; $$
now, since $x_n$ and $x_n'$ converge to the same limit, $\lVert x_n-x_n'\rVert\to 0$. We conclude that $\lVert T_0x_n-T_0x_n'\rVert\to 0$, which immediately implies (2).
Best Answer
For the proof of equivalence, you can find it on Wikipedia.
Since a normed vector space is a metric space with metric induced from the norm, you can just copy the definition of continuity at $x_0$ for real functions of real variable:
$$(\forall\varepsilon >0 )(\exists\delta > 0 )\ \|x-x_0\|<\delta \implies \|Ax-Ax_0\|<\varepsilon.$$
You might be confused by the proof since any function $f$ that satisfies $$\|f(x)-f(y)\|\leq C\| x-y\|$$ for some $C>0$ must be continuous. Try to prove it.
Also, it might be worthwhile for you to try to prove that continuity of linear operator at $0$ implies continuity at all points.