Let's analyze the error.
The figure shows plots of the distribution function of various Bernoulli$(p)$ variables in blue and the corresponding Normal distributions in Red. The shaded regions show where the functions differ appreciably.
(Why plot distribution functions instead of density functions? Because a Bernoulli variable has no density function. The densities of good continuous approximations to Bernoulli distributions have huge spikes in neighborhoods of $0$ and $1.$)
No matter what $p$ may be, for some values of $x$ the difference between the two distribution functions will be large. After all, the Bernoullli distribution function has two leaps in it: it jumps by $1-p$ at $x=0$ and again by $p$ at $x=1.$ The Normal distribution function is going to split the greater of those two leaps into two parts, whence the larger of the two vertical differences--the largest error--must be at least $1/4.$ In fact, it's always greater even than that.
Here is a plot of the maximum difference between the two functions, as it depends on $p:$
It is never smaller than $0.341345,$ attained when $p=1/2.$ Because probabilities all lie between $0$ and $1,$ this is a substantial error. It is difficult to conceive of circumstances where this approximation would be acceptable, except perhaps when $x\lt 0$ or $x\gt 1:$ but then why use a Normal distribution at all? Just approximate those values as $0$ and $1,$ respectively, without any error at all.