Definition of continuity at a point In "Principles of mathematical analysis" Walter Rudin gives the definition of 
$$\lim_{x \rightarrow p} f(x)=q$$
for $f: X \supset E \rightarrow Y$ with $X,Y$ metric spaces, in this way:

for every $\epsilon>0$ there exists a $\delta >0$ such that
$$d_Y(f(x),q)<\epsilon$$
for all points $x \in E$ for which
$$0<d_X(x,p)<\delta$$

where $d_X$ and $d_Y$ are the distances in $X$ and $Y$. After few pages he gives the definition of continuity:

$f$ is said to be continuous at $p$ if for every $\epsilon>0$ there
  exists a $\delta >0$ such that
$$d_Y(f(x),f(p))<\epsilon$$
for all points $x \in E$ for which 
$$d_X(x,p)<\delta$$

My question is, why did he drop the $0<\dots$ in $d_X(x,p)<\delta$? I though that continuity in $p$ means that the limit of the function in $p$ exists and is equal to $f(p)$, so one should simply replace $q$ with $f(p)$ in the above definition of limit, why changing the condition on $x$?
Doing so a function like
$$f: \{1\} \rightarrow \mathbb{R}$$ which maps 1 to 5, for example, would be continous in 1, since $x=p$ is allowed, but what is the limit of $f$ in 1? It doesn't exist, right?
Is there a deep reason for such a counterintuitive definition? 
EDIT:
Maybe I should better clarify my question. I know that this definition works, I am asking why considering a function continuos at isolated points (which is an immediate consequence of this defintion ---> It turned out I was wrong about this).  
The only reason I can think of is that this definition agrees with the topological one for continuity, but it doesn't seem to me a good reason since topology came after (and would have agreed no matter the convention).
 A: The real question is, why is the $0<d_X(x,p)$ condition in the definition of the limit?
Basically, it doesn't hurt to allow $x=p$ in the continuity example, because (1) we know that $p\in E$, and (2), when $x=p$, $d_Y(f(x),f(p))=0<\epsilon$, so there is no reason to leave it out.
Essentially, in the continuity case, the case $x=p$ is trivially true.
On the other hand, when $p\in E$ in the limit definition, we don't want the limit to depend on $f(p)$, but only on the values $F(x)$ when $x\neq p$.
For example, when $f(x)=0$ for $x\neq p$ and $f(p)=1$, we want $\lim_{x\to p} f(x) = 0$, but that would not be true if we didn't have the condition  $0<d_X(x,p)$ - without that condition, the limit is undefined.
Isolated points: 
We consider isolated points to be points of continuity because we want in general that if:
$f:E\to Y$ is continuous at $p$ and $p\in E'\subset E$ then $f_{|E'}:E'\to Y$ to also be continuous at $p$.
However, note that the continuity definition could have said $0<D_X(x,p)$. That doesn't affect the continuity at $p$ one bit, so if you wanted to define all isolated points as points of discontinuity, you'd want some other definition completely.
A: It's fine to let $x=p$ in the definition of continuity, since then $f(x)=f(p)$, and so $$d_Y(f(x),f(p))=0<\epsilon.$$ It just isn't very interesting. When simply discussing limits of functions, though, we may want to examine behavior near but not at a point (such as when taking derivatives), so we require $x\neq p$ in such cases.
Note: It's possible that $p$ is isolated in $X$, so that we can't talk about the limit as $x$ approaches $p$ at all. A function $f:X\to Y$ is continuous at $p\in X$ if and only if either (i) $p$ is isolated in $X$ or (ii) $p$ is a limit point of $X$ and $f(p)=\lim\limits_{x\to p}f(x)$.

Edit: Even with your proposed alteration to the definition, a function is necessarily continuous at isolated points of its domain. Suppose that $f:E\to Y$ with $E\subseteq X$ and $p$ an isolated point of $E$. In other words, there is some $\delta>0$ such that for any $x\in E$, we have $d_X(x,p)<\delta$ if and only if $x=p.$ In particular, there is no $x\in E$ such that $0<d_X(x,p)<\delta$.
Now, take any $\epsilon>0$. Then for all points $x$ in the (empty) set $$\{x\in E: 0<d_X(x,p)<\delta\},$$ we have that $$d_Y\bigl(f(x),f(p)\bigr)<\epsilon,$$ vacuously. Since $\epsilon>0$ was arbitrary, then $f$ is continuous at $p$, under your proposed alteration to the definition. In fact, your altered version is mathematically equivalent to Rudin's version.
This sort of vacuous truth is bothersome to many, and permitting $x=p$ as in Rudin's definition of continuity allows us to avoid such things. As I said, allowing $x=p$ isn't a problem, but it isn't very interesting, either. The primary reason that we have to have $0<...$ in the limit definition, but not necessarily in the continuity definition, is so that we can examine limits of functions with (for example) removable discontinuities at a point, or limits of functions as we approach points at which they are undefined.
A: In the situation where $p$ is an isolated point, then continuity at $p$ is guaranteed no matter what and it doesn't make sense to talk about the limit of $f(x)$ as $x \to p$, so let's suppose that $p$ is not isolated.
A function which is continuous at $p$ in particular has to be defined at $p$. If you read the definition of limit of a function as the input approaches $p$ carefully, you'll notice that it doesn't require the function to be defined in $p$ but only on a set which has $p$ as a limit point.
A: "I though that continuity in p means that the limit of the function in p exists and is equal to f(p), so one should simply replace q with f(p) in the above definition of limit"
Doing so, the definition says for all $\epsilon>0$ there is $\delta>0$ such that if $0<d_X(x,p)<\delta$ then $d_Y(f(x),f(p))<\epsilon$. But note that if $d_X(x,p)=0$ then $x=p$ so $d_Y(f(x),f(p))=0$, so one may trivially drop the "$0<$".
Rudin stipulates $d_X(x,p)>0$ in the definition of limit just so he may say things like $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}=\text{stuff},$$ without having to worry that the fraction isn't defined at $h=0$.
A possibly clearer definition of $\lim_{x\to t} f(x)$ is the common value of $\lim f(x_n)$ for all sequences $x_n$ in $X\backslash\{t\}$ such that $x_n\to t$, provided that these limits exist and are equal.
