Here is an argument that does not require Urysohn's Lemma: fix $x \in X$ and define $f(y)=d(x,y)$. Then $f:X \to [0,\infty)$ is continuous, $f(x)=0$ and $f(y) >0$ for some $y$. Since the range is an interval it is uncountable and so is $X$.
Here is an example (simpler than the one I mentioned in the comments) of a space which is completely regular but not normal.
The Sorgenfrey line is the space $\Bbb R_\ell=(\Bbb R,\tau)$ where a basis for $\tau$ is given by $\{[a,b)\mid a,b\in\Bbb R\}$, this is a completely regular space (actually it is completely normal as well).
The Sorgenfrey plane is the space $\Bbb R_s=\Bbb R_\ell\times\Bbb R_\ell$ with the product topology, or in other words $\Bbb R^2$ where a basis of the topology is given by the half open squares. This space is still completely regular, as the product of two completely regular spaces is completely regular (easy exercise).
Note that the set $\Delta=\{(x,-x\}\mid x\in\Bbb R\}$ is an uncountable closed discrete subspace of $\Bbb R_s$ (draw a picture to convince yourself of this fact) and that this space is separable, since $\Bbb Q^2$ is a dense subset (every open set contains a Euclidean open set, so every open set intersects $\Bbb Q^2$).
Now if $\Bbb R_s$ were normal, then we could use Tietze's extension theorem to show that there are at least $2^{2^{\aleph_0}}$ continuous functions $\Bbb R_s\to\Bbb R$, since every function $\Delta\to\Bbb R$ is continuous and there are $2^{2^{\aleph_0}}$ distinct such functions, all of which could be extended to a function $\Bbb R_s\to\Bbb R$.
But $\Bbb R_s$ is separable, so a continuous function is determined by its values on $\Bbb Q^2$ and there's only $2^{\aleph_0}$ functions $\Bbb Q^2\to\Bbb R$, contradicting the previous paragraph. Note that this argument really shows that a separable space with an uncountable closed discrete subset cannot be normal (so you can also use it to show that the Moore plane, another example of a space which is completely regular but not normal, is in fact not normal).
Let me finish by mentioning two useful resources for this kind of questions: $\pi$-base, an online database of topological spaces, searchable by properties, and the book "counterexamples in topology" by Steen and Seebach, where $\Bbb R_\ell$ and $\Bbb R_s$ are the spaces 51 and 84 respectively.
Best Answer
The first proof is quite correct: it's easier even to say that $f[X] = [0,1]$ (just use $a = 0, b= 1$, as in the standard formulation of Urysohn), as the only connected subset of $[0,1]$ that contains $0$ and $1$ equals $[0,1]$ (if it missed $a \in (0,1)$ we'd have an immediate disconnection). And $|X| \ge |f[X]|$ etc.
Suppose $X$ is countable. Then $X$ is Lindelöf (trivially) and a regular Lindelöf space is normal (see this answer, e.g., or search for other proofs if your text does not cover this). Now by the previous part, $X$ is uncountable, contradiction.