Adjoint to forgetful functor from $\textbf{Met}$ to $\textbf{Set}$ As part of an assignment we need to prove that the forgetful functor $U:\textbf{Met}\to\textbf{Set}$ doesn't have a left adjoint (morphisms in $\textbf{Met}$ are contractions). 
For this, I have tried to show that arbitrary sets need not have a free object in $\textbf{Met}$.  I have a feeling that as soon as you have more than 1 object in a set, you cannot get the free object; ideally I would like to have this bare-minimum counterexample. 
So I have tried multiple ways to prove that $X=\{0,1\}$ doesn't have a free object, but I never really get a general enough result, if that makes sense.  I.e. I make assumptions along the way or deal with just specific examples. 
E.g. I have tried to give  $X$ a metric and have it act as the free object, and that failed the universal property with a specifically chosen test object (since the inclusion map wasn't a contraction).  But this only proves that the free object can't have $X$ as its underlying set; no reason it couldn't be a completely different metric space.  
I have also tried to smartly choose a test for the UP (including $X$ with a variety of metrics, 3-element sets, etc) to no avail, never really being as generic as possible. 
Is there some "easy" counterexample I can use to prove not all sets have a free object in $\textbf{Met}$?

I do know about solution sets, but for now I would rather avoid that topic, as this is an exercise we should be able to do without that or the adjoint functor theorem. 

Let $X = \{0,1\}$ and denote by $X_d$ the metric space with underlying set $X$ and $0$, $1$ having distance $d > 0$. Suppose there exists a free object $F(X)$ of $X$ in $\bf{Met}$, then for every $d > 0$ we get 
$$\text{Hom}_{\bf{Met}}(F(X),X_d) \cong \text{Hom}_{\bf{Set}}(X,X)$$ which consists of four elements. This implies that $F(X)$ decomposes as $F(X) = U_1 \cup U_2$ with $U_1$ and $U_2$ the connected components of $F(X)$. Indeed, $F(X)$ is clearly disconnected and if $F(X)$ were to decompose as disjoint union of open subsets $U_1$, $U_2$ and $U_3$, an appropriate choice of $d$ (smaller than the minimal distances between elements of $U_1$, $U_2$ and $U_3$) would give us more than four maps $F(X) \rightarrow X_d$. 
Now, let $x_i \in U_i$ for $i = 1,2$ and denote by $c$ the distance between $x_1$ and $x_2$. For $d > c$ we get the contradiction that there is a contraction $F(X) \rightarrow X_d$ mapping $x_1$ to $0$ and $x_2$ to $1$.
