I am reading a proof that a certain subset $T^P$ of functions in $C([0,1],\mathbb{R})$ satisfying certain property $P$ is dense. The argument goes as follows: take $f\in T^P$, take another arbitrary $h\in C([0,1],\mathbb{R})$ and make a new function
$$(1-\varepsilon) f+ \varepsilon h$$
show that it satisfies the property $P$ for any $1>\varepsilon>0$. Conclude that $T^P$ is dense in $C([0,1],\mathbb{R})$.
My question is, I can do the same with another pair of functions, say $\bar{f}$ and $\bar{h}$, both failing property $P$ and their convex combination $$(1-\varepsilon) \bar{f}+ \varepsilon \bar{h}$$ also fails $P$. But so then I have constructed an open set in $C([0,1],\mathbb{R})$ that fails $P$ and so $P$ is not dense?
The set you have constructed is not open.
For a simpler apparition of the same phenomenon, consider the subset $$A=\{(x,y)\in\mathbb R^2\mid x\ne0\}\subset \mathbb R^2\,.$$
Then $A$ is dense in $\mathbb R^2$ (with the same proof as the one you gave).
Your argument is that if $(0,y_1)$ and $(0,y_2)$ are not in $A$, then $$\epsilon (0,y_1) + (1-\epsilon)(0, y_2) = (0, \epsilon y_1 +(1-\epsilon) y_2)$$ is not in $A$, which is true.
However the set $$B = \{(0, \epsilon y_1 +(1-\epsilon) y_2)\mid 0\lt\epsilon\lt1\}$$ is not open. Indeed, if is was, then for some $\epsilon>0$, $(\epsilon,\frac{y_1+y_2}{2})$ would be in $B$, but it clearly isn't.