$C^1[a,b]$ is closed in $C[a,b]$

Try building a sequence of differentiable functions, converging to $f(x) = |x|$ (say $[a, b] = [-1, 1]$).

No -- on the contrary, $C^1[a,b]$ is dense in $C[a,b]$, by the Weierstrass approximation theorem. But that's too big a gun to bring out to show that it is simply not closed.

Instead, to follow your ansatz, Karolis' suggestion is to consider $h(x)=|x|$. That's the archetypical example of a function that's continuous but not differentiable, if $0$ is an interior point of $[a,b]$ (which we can clearly assume without loss of generality).

We're then looking for an $\epsilon$ such that no $f$ that differs from $h$ by at most $\epsilon$ can be differentiable. If we fail in that -- in other words, if for every $\epsilon$ there is a differentiable $f$ close to $h$ -- then we have proved that $C^1[a,b]$ is not closed.

Now consider $f(x)=\sqrt{x^2+\delta}$ where $\delta$ is a small positive constant that possibly depends on $\epsilon$. Is this always differentiable? Can we made it stay within a distance of $\epsilon$ from $|x|$ by choosing $\delta$ small enough?