Your derivation doesn't give the correct result because the limit $\lim_{t\to\infty}e^{-j\omega t}$ doesn't exist. You seem to have confused that limit with $\lim_{t\to\infty}e^{-\omega t}$, which does exist and which equals zero for $\omega>0$.
There are many ways to derive the Fourier transform of the unit step function $u(t)$ (see e.g., this answer), but I'll show you one which doesn't involve more than just engineering math. At least if you accept some intuitively pleasing facts such as that the derivative of the unit step equals the Dirac delta impulse:
$$u'(t)=\delta(t)\tag{1}$$
It's important to note that $(1)$ holds for any function with a discontinuity at $t=0$ that is otherwise constant, satisfying $u(0^+)-u(0^-)=1$. Hence, $(1)$ only defines $u(t)$ up to a constant. Taking the Fourier transform of $(1)$ gives
$$j\omega U(\omega)=1\tag{2}$$
It would be tempting to conclude from $(2)$ that $U(\omega)=1/j\omega$ holds, however, this is not correct. The reason is the additive constant that got lost due the derivative in $(1)$.
Hence, the correct conclusion from $(1)$ is
$$U(\omega)=c\delta(\omega)+\frac{1}{j\omega}\tag{3}$$
where $c$ is the constant that can be added to any function satisfying $(1)$ without violating Eq. $(1)$. We just have to find the specific value of $c$ for which $(3)$ holds, with $U(\omega)$ being the Fourier transform of the unit step.
It's easy to see for which function $c=0$ holds. Since $1/j\omega$ is purely imaginary, we know that the corresponding time domain function is odd, it has a jump of height $1$ at $t=0$, and it is otherwise constant. The only function satisfying all of these requirements is a scaled signum function, i.e.,
$$\frac12\textrm{sgn}(t)\Longleftrightarrow \frac{1}{j\omega}\tag{4}$$
Since the unit step function satisfies
$$u(t)=\frac12+\frac12\textrm{sgn}(t)\tag{5}$$
we finally arrive at the Fourier transform of $u(t)$ by combining $(4)$ and $(5)$:
$$\bbox[#f8f1ea, 0.6em, border: 0.15em solid #fd8105]{
U(\omega)=\pi\delta(\omega)+\frac{1}{j\omega}}\tag{6}$$
