Can we define a metric on $Y$ such that all continuous mappings $f:X\rightarrow Y$ are constant? Given that $Y$ contains more than one element and let $X$ be the real line equipped with the standard metric. Then can we define a metric $\sigma$ on $Y$ such that every continuous mapping $f:X\rightarrow Y$ is constant?
Now I believe the answer to this is yes and with $\sigma$ being the discrete metric. Here's my reasoning... If $f$ is not constant then $\sigma(f(x),f(y))=1$ for all $f(x)\neq f(y)$. However as $f$ is continuous we must have that $\sigma(f(x),f(y))\leq \epsilon$ whenever $|x-y|<\delta$. Choosing $\epsilon <1$ achieves a contradiction and thus we must have that $f$ is in fact constant. 
Is this a good enough argument?
 A: Your argument doesn't distinguish between continuity and uniform continuity.  With uniform continuity, the $\delta$ does not depend on the $x$.
For example, $x\mapsto e^x$ is a continuous function, but given $\varepsilon>0$ it is not true that there exists $\delta>0$ such that for all $x,y\in\mathbb R$, if $|x-y|<\delta$ then $|e^x-e^y|<\varepsilon$.  One can always make $x$ and $y$ so big that $|e^x-e^y|>1$ for some $x,y$ for which $|x-y|<\delta$.  I.e. the exponential function is not uniformly continuous.
But for each $x$ separately, it is true that that for all $\varepsilon>0$, there exists $\delta>0$ such that for all $y\in\mathbb R,$ if $|x-y|<\delta$ then $|e^x-e^y|<\varepsilon$.
In other words, the value of $\delta$ can depend on $x$.  There is no $\delta$ that simultaneously works for all values of $x$, but for each $x$ separately there is a sufficiently small $\delta>0$.
You need to rephrase your argument to allow $\delta$ to depend on $x$.
Thus suppose $x\in\mathbb R$ and $\varepsilon\in(0,1)$.  Since $f$ is continuous at $x$, there is some $\delta>0$  (possibly depending on $x$) such that if $|x-y|<\delta$ then $\sigma(f(x), f(y))<\varepsilon<1$.  The nature of the metric is such that if $\sigma(u,v)<1$ then $u=v$.
This shows $f$ is constant on every sufficiently small neighborhood of $x$, where sufficiently small means the radius is $<\delta$.
A difficulty is that how small is sufficiently small has not been shown to be independent of $x$.  If it were, then for any $x,y\in\mathbb R$, no matter how far they are from each other, we could just lay some overlapping open intervals of length $\delta$ down on the line and with enough of them they'd reach from $x$ to $y$, and we'd have $f(x)=f(y)$.
Ultimately your argument shows $f$ is constant on each connected component of $\mathbb R$.  To complete this argument, you need to cite or prove the fact that $\mathbb R$ has only one connected component.  So the task is somewhat more onerous than it at first appears.
A: Theorem: Let $X$ be a connected topological space and let $Y$ be a set equipped with the discrete metric $d$, i.e. $d(a,b)=1$ if $a\neq b$ and $d(a,a)=0$ for all $a\in Y$. Then a function $f:X\to Y$ is continuous if and only if $f$ is constant.
Proof: The proof of "any constant function is continuous" is straight forward (the inverse image of any open set is empty or $X$).
Now to prove the converse let $f:X\to Y$ be continuous. For every $x\in X$ there exist an open set $V_x\subset X$ such that $x\in V_x$ and $d(f(x),f(y))<\frac{1}{2}$ for all $y\in V_x$. Since $d$ is the discrete metric on $Y$ we conclude that $f$ is constant on $V_x$.
Now we will show that if $f$ is not constant on $X$ then $X$ cannot be connected leading to a contradiction. Let $x_0\in X$.  Consider 
$$A:=\bigcup_{{x\in X \atop f(x)=f(x_0)}}V_x \,\,\mbox{ and }\, B:=\bigcup_{{x\in X \atop f(x)\neq f(x_0)}} V_x.$$
Notice that $A$ and $B$ are disjoint open subspaces of $X$ such that $X=A\cup B$ and $x_0\in A$. If $f$ is not constant, then there exist $y\in X$ such that $f(y)\neq f(x_0)$. Then $y\in V_y\subset B$. Hence $B$ is non-empty and therefore $X$ is not connected, a contradiction. $\blacksquare$
