There exists continuous, nowhere differentiable functions that are, by the Stone-Weierstrass theorem, approximated by polynomials. So, a sequence of polynomials, which are obviously $C^1$ functions, are converging to a function that is not $C^1,$ but $C^0$. So, how is it that $C^1$ is a Banach space?
I'm not asking someone to prove that $C^1$ is complete at all, I'm asking why the conditions I specified do not directly contradict the completeness.
I will condense the discussion in the comments into an answer. Scroll to the end if you want the actual mechanical explanation. The rest is a general explanation of why we shouldn't expect spaces with different norms to act the same.
First things first, we have to speak about a few general observations and facts.
A metric space $(X,d)$ is a set $X$ with a metric $d$. In particular, the same set $X$ with different metrics $d_1$ and $d_2$ defines different metric spaces. They may not even share a single relevant feature except their cardinality. The circle, the disc, the Euclidean plane, the line, and Euclidean space are all completely different metric spaces. But all of these spaces can be defined using the same set, since they have equal cardinalities. Take a bijection $f:\mathbb R\to M$, where $(M,d)$ is any of those metric spaces, then the metric space $(\mathbb R,f(d))$ is isomorphic to $(M,d)$ as a metric space, meaning that it has essentially the same structure. So we can define completely different metric spaces on the exact same set using different metrics. And you may already notice: some of these are complete (the circle, the line, the plane, 3d space), while others are not (the disc). This is because the underlying set has no influence on the metric space except supplying a cardinality. But metric spaces of equal cardinality can behave wildly differently.
Your question deals with normed spaces $(V,\Vert\cdot\Vert)$ instead of metric spaces, but for any norm $\Vert\cdot\Vert$ the function $d(x,y):=\Vert y-x\Vert$ is a metric, and so the same thing as above applies to normed spaces, too: a different norm on the same vector space may lead to completely different normed spaces. The only dangerous thing is that this doesn't apply to finite dimensional normed spaces, so we may be lulled into a false sense of security while using only finite dimensional spaces, to then be surprised when everything goes back to normal in infinite dimensional spaces.
Anyway, we deal with the following infinite dimensional normed spaces: $(C^0,\Vert\cdot\Vert_0),(C^1,\Vert\cdot\Vert_0),(C^1,\Vert\cdot\Vert_1)$, where I always shorten $C^n([a,b])$ to $C^n$. The norms are given by $\Vert f\Vert_0:=\sup_{x\in[a,b]}\vert f(x)\vert$ and $\Vert f\Vert_1:=\Vert f\Vert_0+\Vert f'\Vert_0$.
Now Stone-Weierstraß says that in $(C^0,\Vert\cdot\Vert_0)$ the subspace $(P,\Vert\cdot\Vert_0)$ of polynomial functions is dense. Since $(P,\Vert\cdot\Vert_0)<(C^1,\Vert\cdot\Vert_0)<(C^1,\Vert\cdot\Vert_0)$, where $<$ means "is subspace of", we also get that $(C^1,\Vert\cdot\Vert_0)$, as an intermediate space between a subspace and its completion, must also be dense in the superspace, and thus not complete (since complete spaces are never dense in proper superspaces). This is what you observed.
However, $(C^1,\Vert\cdot\Vert_1)$ is not an intermediate space between $(P,\Vert\cdot\Vert_0)$ and $(C^0,\Vert\cdot\Vert_0)$. It doesn't even have the same norm! It may behave completely differently to $(C^1,\Vert\cdot\Vert_0)$. And thus the theorem that $(C^1,\Vert\cdot\Vert_1)$ is complete has no bearing on the observations about $(C^1,\Vert\cdot\Vert_0)$.
But you wanted to know about a mechanism that suppresses convergence of polynomials to non-smooth functions in the norm $\Vert\cdot\Vert_1$. So I'll give you that as well, though without proof. Convergence of a sequence $f_n$ in $ \Vert\cdot\Vert_0$ means that starting at some $n$, $f_n$ and the limit function $f$ are close to each other on their whole domain. This is because $\Vert f-f_n\Vert_0$ measures the largest difference between $f$ and $f_n$ on their whole domain. Now if this is the only requirement, then polynomial sequences can converge to non-smooth functions. But if we look at the derivatives $f'_n$ of such a sequence, we will notice that this new sequence $f'_n$ will not converge in $\Vert\cdot\Vert_0$. It's not even Cauchy. So to forbid sequences of polynomials from converging to non-smooth functions, we can erect an additional barrier and introduce a new kind of convergence ($C^1$-convergence) in which we require both $f_n$ and $f'_n$ to converge in the old sense ($C^0$-convergence). We can force this by introducing the norm $\Vert f\Vert_1$, which is the sum of the $C_0$ norms of $f$ and $f'$. So if this norm is small, then both the function and its derivative are small, which is a stricter requirement than only having the function itself be small. And a Cauchy sequence in this norm will require not only the functions themselves to get ever closer to each other, but their derivatives, too! It's this last bit that stops convergence to non-smooth functions.
