If $f$ is continuous on $\mathbb{R}$, then it is continuous on a closed interval $[a,b]\subset \mathbb{R}$. I am writing a proof for a problem, and one of the ways to proceed would be using the following claim:

If  $f$ is continuous on $\mathbb{R}$, then it is continuous on a closed interval $[a,b]\subset \mathbb{R}$.

I think it is true since the definition of continuity states that if $f$ is continuous on a set $S$, then  $\forall x_0 \in S$ and $\epsilon>0$ there is some $\delta>0$, such that $x \in S$ and $|x-x_0|<\delta$ imply $|f(x)-f(x_0)|<\epsilon$; so the points coming out of the subset of $S$, must be from $S$, and thus the existence of $\delta$ for the whole set implies an existence of $\delta$ for any of its subsets. 
However, if $f$ is continuous on any closed interval $[a,b]\subset \mathbb{R}$., then it must be uniformly continuous on any such interval, which in turn would imply it is uniformly continuous on $\mathbb{R}$. But this would mean that the continuity of a function implies its uniform continuity, which is clearly not true. So there must be a flaw in my argument, which I cannot see. (I am expecting it to be something like "a function being uniformly continuous on any $[a,b]$ closed interval does not imply it is continuous on $\mathbb{R}$," which I don't understand why would be the case, since you can take the $\delta$ of the uniform continuity to be the minimum of all deltas across the intervals, no?)
EDIT: There seems to be confusion regarding my question, so let me narrow it down. 
Is this true: If for all $a,b \in \mathbb{R}$, the function $f$ is uniformly continuous on $ [a,b]\subset \mathbb{R}$, then $f$ is uniformly continuous on $\mathbb{R}$.
 A: 
which I don't understand why would be the case, since you can take the $\delta$ of the uniform continuity to be the minimum of all deltas across the intervals, no?

The minimum might be $0$, which is not allowed. For example:
Let $f(x)=x^2$. Then on $[-b,b]$ we have that $|f(x)-f(y)|=|x^2-y^2|=|x-y||x+y|\leq 2b |x-y|$.  
So given an $\varepsilon>0$, you can let $\delta<\varepsilon/(2b)$. Thus $f$ is uniformly continuous on $[-b,b]$.
The infimum over $b$ of this is exactly $0$, which is not allowed. 
A: It simply doesn't follow if if a condition must be true for any and every $[a,b]$ (based on the value of $a,b$) that it will  be true for all $\mathbb R$ regardless of the what interval we are looking at.
A trivial counter-example would be that every $[a,b]$ has a maximal element, $b$, but clearly  $\mathbb R$ does not.  The argument breaks down, in that the condition, having a maximal element, is very dependent upon which closed interval we choose.
One can argue:  Well if $[a,b]$ has a maximal element and $[a_i, b_i]$ has maximal element then $[a,b]\cup [a_i,b_i]$ will have a maximal element, $\max\{\max[a,b],\max[a_i,b_i]\}$, then $\mathbb R$ will have a maximum element that is equal to $\max\{\max[a,b]| a, b\in \mathbb R\}$.  
The problem there is that $\mathbb R$ is unbounded and $\max\{\max[a,b]| a, b\in \mathbb R\}$ simply does not exist.
Uniform continuity is very similar.  To be uniformly continuous on closed interval $[a,b]$ means for any $\epsilon > 0$ there is a $\delta > 0$ so that whenever $|x-y| < \delta$ AND whenever $a\le x \le b, a\le y\le b$ then $|f(x)-f(y)|< \epsilon$.  The condition of $x,y\in [a,b]$ can not be ignored.
You might argue that for an $\epsilon$ there is $\delta_{a,b}$ so that... $|x-y]<\delta_{a,b}\implies |f(x)-f(y)| < \epsilon$ for all $x,y \in [a,b]$... then if we take $\delta = \min(\delta_{a,b}) > 0$, then if $|x-y|< \delta$ then for any $[a,b]$ so that $x,y\in [a,b]$ then $|x-y|< \delta \le \delta_{a,b}\implies |f(x)-f(y)| < \epsilon$.
And this would but true if $\delta = \min(\delta_{a,b}) > 0$ existed. 
It'd even be true if $\delta = \inf(\delta_{a,b}) > 0$ existed but $\min$ did not.
However as $\inf(\delta_{a,b})$ could be $0$ the argument fails in those cases as $|x-y|< \delta =0$ is impossible.
The classic counter-example if $f(x) = x^2$.  For any $[a,b]$ and $\epsilon > 0$ then if  $\delta_{a,b} = \frac {\epsilon}{2\max(|a|,|b|)}$ then $|x-y|< \delta$ would imply $|x^2 - y^2|=|x-y||x+y| \le |x-y|*(2\max(|a|,|b|)) < \delta_{a,b}*2\max(|a|,|b|)=\epsilon$.
But $\inf \delta_{a,b} = 0$ so the argument fails.
(And the proof that $f(x)=x^2$ is not uniformly continuous is standard: For any $\epsilon> 0$, let $\delta > 0$ and let $x_1$ be so that $x_1\frac {\delta}2 + \frac {\delta^2}{4}>\epsilon$ (i.e. $x_1 > (\epsilon - \frac {\delta^2}4)\frac 2{\delta}$) then for any $x,y\in (x_1-\frac \delta 2, x_1+\frac \delta 2)$ we have $|x_1 - (x_1+\frac {\delta}2)|=\frac {\delta}2 < \delta$ while $|(x_1+\frac {\delta}2)^2 - x_1^2| = x_1\frac {\delta}2 + \frac {\delta^2}{4}>\epsilon$.)
