Consider the set $C^1(\mathbb{R},\mathbb{R})$ of continuously differentiable functions on $\mathbb{R}$, endowed with the extended $C^1$ norm $\|f\|_{C^1} = \sup_{x\in \mathbb{R}} |f(x)| + \sup_{x\in \mathbb{R}}|f'(x)|$.
If we proceed in the usual way (as if the extended norm were a norm, inducing a metric, inducing a topology), do we get a Baire space?
If I understand the question the answer is yes.
I take it we're talking about an "extended metric", by which I gather we mean a function $d:X\times X\to[0,\infty]$ that satisfies all the axioms for a metric. Assuming that's what we mean, the reason people don't talk about extended metric spaces is that an extended metric space is a metric space. If $d$ is an extended metric on $X$ and we define $$\tilde d(x,y)=\min(1,d(x,y))$$then $\tilde d$ is a metric on $X$ that induces the same topology as $d$.
In particular $C^1$ with your extended norm is a complete metric space. Depending on why you care about this it may or may not be important to point out that $C^1$ with this (extended) metric is not a topological vector space. Addition is continuous, but scalar multiplication is not. (Say $f_n(t)=t/n$. Then $d(f_n,0)\not\to0$ as $n\to\infty$.)
Again, depending on why you're asking the question, it may or may not be a good idea to point out that $C^1(\Bbb R)$ is a topological vector space in a natural and useful way that has little to do with that extended norm. Define $$\rho_n(f)=\sup_{|t|\le n}(|f(t)|+|f'(t)|)$$and $$d(f,g)=\sum_{n=1}^\infty2^{-n}\min(1,\rho_n(f-g)).$$Then $d$ is a complete metric on $C^1$ making it into a topological vector space. And we have $d(f_n,f)\to0$ if and only if $f_n\to f$ and $f_n'\to f'$ uniformly on compact sets, showing that this topology captures the natural notion of convergence in $C^1$.