For the first question, following your thoughts, let $f$ be holomorphic(i.e. analytic) on $D_1(0)$, such that ${\rm Re}f=u$, i.e. $u$ is the real part of $f$. Then the Taylor expansion of $f$
$$f(z)=\sum_{k=0}^\infty a_kz^k$$
converges to $f$ uniformly on compact subsets of $D_1(0)$(and hence so do the corresponding real and imaginary parts). For every $n\ge 0$, let
$$u_n(z)={\rm Re}\big(\sum_{k=0}^n a_kz^k \big).$$
Since $u_n$ is the real part of a (holomorphic) polynomial of $z=x+iy$, $u_n$ is a real valued harmonic polynomial of $x$ and $y$. Moreover, $u_n$ converges to the real part of $f$, which is just $u$, uniformly on compact subsets of $D_1(0)$.
For the second question, assume that there is a sequence of real valued harmonic polynomials $(u_n)$, such that it converges to $\ln|z|$ uniformly on compact subsets of $\Omega$. Note that $u_n$ is in fact harmonic on $D_2(0)$, so by maximum/minimum principle, it is easy to see that $u_n$ converges uniformly on compact subsets of $D_2(0)$, and let us denote the limit by $u$. Then using Poisson integral formula or otherwise, it is easy to see that $u$ is harmonic on $D_2(0)$. However, since $\ln|z|$ is not harmonic at $0$, it contradicts to the fact that $u=\ln|z|$ on $\Omega$, so such $(u_n)$ does not exist.
I'd like to attempt an answer. I will have no problem with down-votes or with deleting it, since I am not 100% confident in it.
Let $A$ be the annulus $\frac{2}{3} \le |z| \le \frac{3}{2}$. For a sequence $a_n \in \mathbb{C}$, define $f_{a_n}(z) = f(a_n z)$ and $B_{a_n} = \{a_n z : z \in A \}$.
$B_{a_n}$ is an annulus with inner radius $|a_n|\frac{2}{3}$ and outer radius $|a_n|\frac{3}{2}$.
Let $a_n = n$, n positive integer. Let $f_{n_k}$ be the subsequence of $f_n$ guaranteed by the statement. Then either $f_{n_k} \to \infty$ uniformly on $A$ or $f_{n_k} \to g$ uniformly on $A$ for an analytic function $g$.
In the second case, $g$ is analytic so $g(A)$ is bounded. Using uniform convergence, you can see that There is an $M$ such that $f_{n_k}(A) < M$ for all $k$ sufficiently large. This means $f(z) < M$ for all $z \in B_{n_k}$, with $k$ sufficiently large. Mimicking the proof of Liouville's theorem that uses the Cauchy Integral Formula, this shows that $f(z)$ is constant.
In the first case, for any $M$ you can achieve $|f_{n_k}(z)| > M$ for $k$ sufficiently large. If in addition, for $k$ sufficiently large, $B_{n_k} \cap B_{n_{k+1}}$ is non-emtpy, then this shows $f(z) \to \infty$ as $z \to \infty$, and therefore $f$ must be a polynomial.
If on the other hand there are infinitely many $k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, then we need to look at what happens in these "gaps". Note that the outer radius of $B_{n_k}$ is in this case strictly less than the inner radius of $B_{n_{k+1}}$. This "gap" is an open annulus, but $f$ is analytic on the closure. Suppose such a gap does not contain a zero of $f$. Then the Minimum Modulus Theorem says that the modulus of $f$ in the gap is at least as great as the minimum modulus on the boundary, and so in the gap $f|(z)| > M$.
Therefore if only finitely many gaps contain zeros, we again see that $f(z) \to \infty$ as $z \to \infty$.
Finally, suppose there are infinitely $n_k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, and infinitely many of these "gaps" contain a zero of $f$. For the $m$-th gap pick a zero $b_m$, with $|b_m| < |b_{m+1}|$ and $|b_m| \to \infty$. These $b_m$ define a new sequence $f_{b_m}$ of functions on $A$, a subsequence of which converges uniformly. But in this case it cannot converge uniformly to $\infty$, because each function in the sequence has a zero at $z = 1$. Then as in the proof of the second case, $f$ must be constant.
Best Answer
It seems that $u + iv$ must be analytic. The harmonicity of $u$ and $v$ ensures that $f= u_x - iu_y$,$ g= v_y + iv_x$ are analytic on $\Omega$ (one can check that these two functions satisfy the C-R equations). Since $f$ and $g$ agree on a set with a limit point, they must agree on all of $\Omega$, hence $u$ and $v$ satisfy the C-R equations on all of $\Omega$. So, indeed, $u + iv$ is analytic on all of $\Omega$.