Consider the normed space $C([0, 1])=\{f:[0, 1]\to\mathbb{R}\mid f\text{ is continuous}\}$ with the sup norm $$\|f\|_{\infty}=\max_{x\in [0,1]}|f(x)|.$$ Let $$B=\left\{f\in C([0, 1])\;\middle|\; \int_{0}^{1}f(x)^2\,dx<1\right\}.$$ Is $B$ connected?
I only proved that $B$ is an open subset of $C([0, 1])$.
It is sufficient to prove that $B$ is path-connected, because path-connectedness implies connectedness.
Suppose that $f,g$ are two elements of $B$, and define the function $p$ from $[0,1]$ to $B$ by: $$p(s) := (1-s)f + sg.$$ Clearly, $p(0) = f$ and $p(1) = g$. In addition $p(s) \in B$, because a linear combination of continuous functions is continuous and $$ \int_{0}^{1} p(s)(x)^2 \, dx \leq \int_0^1 (1-s)f(x)^2 + sg(x)^2 \leq 1 \, dx.$$ Here we used the convexity of the square function.
It remains to prove that $p(\cdot)$ is continuous. Let us assmue that $s_n \to s$ in $[0,1]$. Because $p(s) - p(s_n) = (s-s_n)(g-f),$ we see that $$\|p(s) - p(s_n)\|_{\infty} = (s-s_n) \|g -f \|_{\infty} \to 0,$$ which concludes the argument.