Is every connected metric space with at least two points uncountable? [duplicate]

As the topic, how to prove that every connected metric space with at least two points uncountable? Of course i know the definition that a countable set mean there is a bijection between the set and the positive integer. Connected is opposite of disconnected where the set can partition into two disjoint open sets.

Let us have another proof.

Since $X$ has at least two elements, let us choose $x_0,x_1\in X$, $x_0\neq x_1$. Define $f:X\rightarrow [0,1]$ by $$ f(x):=\frac{d(x,x_0)}{d(x,x_0)+d(x,x_1)},\text{ for all }x\in X. $$ Clearly, $f$ is continuous and $$f(x_0)=0\text{ and }f(x_1)=1.$$ Since $X$ is connected and the continuous image of connected space is connected (so called intermediate value theorem), it follows that $$ f(X)=[0,1], $$ which shows that $X$ is uncountable because $[0,1]$ is uncountable. This proves the result.

Not only must $X$ be uncountable, its cardinality must be at least $2^\omega=\mathfrak c$.

Let $\langle X,d\rangle$ be a metric space, and suppose that $|X|<2^\omega$. Fix $x,y\in X$ with $x\ne y$, and let $r=d(x,y)>0$. Let $D=\big\{d(x,z):z\in X\big\}$; $|D|\le|X|<2^\omega=|(0,r)|$, so there is a real number $s\in(0,r)\setminus D$. Show that $B_d(x,s)$ is a non-empty clopen subset of $X$ whose complement is also non-empty, and conclude that $X$ is not connected.

This is perhaps a generalisation of the method of user49521's answer above. Suppose now that $X$ is no longer a metric space but a normal space with at least two points that is connected. Call those two points $x_0$ and $x_1$. Then the Urysohn Lemma gives the existence of a continuous function $f : X \to [0,1]$ such that $f(x_0) = 0$, $f(x_1) = 1$. Now because $X$ is connected its image is also connected. Connected subsets of the reals are intervals and so we conclude that $X$ surjects via $f$ onto some interval that has cardinality $\mathfrak{c}$, so that $|X| \geq \mathfrak{c}$.