We can use your second estimate
$$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + R(x)$$
where $\displaystyle R(x) = \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$
to show the asymptotic estimate:
$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \mathcal{\theta}\left(\frac{\sqrt{x}}{\log x}\right)$$
In fact something stronger is known, that
$\displaystyle R(x) = \mathcal{O}\left(\frac{1}{\log^2{x}}\right)$
and I believe this gives us
$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log ^2x}\right)$$
For this, we use Abel's Identity.
Define $\displaystyle A(x) = \sum\limits_{n \leq x} \frac{a(n)}{n}$, where $\displaystyle a(n) = 1$ iff $\displaystyle n$ is prime and $\displaystyle 0$ otherwise. The above estimate tells us that $\displaystyle A(x) = \log\log x + A + R(x)$.
Setting $\displaystyle f(t) = \sqrt{t}$ and using Abel's Identity we obtain
$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} + C = A(x) \sqrt{x} - \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t$$
Now
$$\displaystyle \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t = \int_{3}^{x} \frac{\log \log t + A + R(t)}{2\sqrt{t}} \ \text{d}t$$
$$\displaystyle = C^{\prime} + A\sqrt{x} + \sqrt{x} \log \log x - \mathrm{Ei}\left(\frac{\log x}{2}\right) + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right)$$
using
$$\displaystyle \int \frac{\log \log x }{\sqrt{x}} = 2\sqrt{x} \log \log x - 2\mathrm{Ei}\left(\frac{\log x}{2}\right) + K$$
and
$$\displaystyle \int \frac{R(x)}{\sqrt{x}} = \mathcal{O}\left(\int \frac1{\log^2 x \sqrt{x}}\right) = \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$
where $\displaystyle \mathrm{Ei}(x)$ is the exponential integral.
(All gotten using Wolfram Alpha).
Since $\displaystyle \mathrm{Ei}(x) = \frac{e^x}{x} + \mathcal{O}\left(\frac{e^x}{x^2}\right)$ (again using Wolfram Alpha), we get
$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} = C + R(x) \sqrt{x} + \frac{2\sqrt{x}}{\log{x}} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$
(Hopefully, I got the computations right).
I will leave my earlier answer here (which is more relevant to the question as asked)
An elementary lower bound
Here is one elementary way (which does not use your second estimate or the prime number theorem) which proves something slightly stronger.
Using the fact the each integer is uniquely expressible as the product of a square and a square free integer, we have that
$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \prod_{p \le n} \left(1 + \dfrac1{\sqrt{p}}\right) \sum_{k=1}^{n} \dfrac1{k}$$
Now $\displaystyle e^x \gt 1 + x$, we have that
$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \left(\prod_{p \le n} e^{\dfrac1{\sqrt{p}}}\right)\left(\sum_{k=1}^{n} \dfrac1{k}\right)$$
We use the estimate that
$$\displaystyle \sum_{j=1}^{n} \dfrac{1}{\sqrt{j}} \ge 2\sqrt{n-1}-2$$
and that
$$\displaystyle \sum_{k=1}^{n} \dfrac1{j} \le \log (n+2)$$
to get
$$\displaystyle 2\sqrt{n-1}-2 \leq \log(n+2) \prod_{p \le n} e^{\dfrac1{\sqrt{p}}}$$
Taking logs, we see that
$$\displaystyle \log (\sqrt{n-1}-1) - \log \log(n+2) + \log 2 \le \sum_{p \le n} \dfrac1{\sqrt{p}}$$
Since, $\displaystyle \log(\sqrt{n-1}) - \log (\sqrt{n-1} -1) \to 0$ as $\displaystyle n \to \infty$,
$\displaystyle \log \log (n+2) - \log \log n \to 0 \ \text{as} \ n \to \infty$
$\displaystyle \log (n) - \log (n-1) \to 0 \ \text{as} \ n \to \infty$ and
$\displaystyle \log 2 \gt 0$, you just need to verify you inequality for finite number of $\displaystyle n$.
Best Answer
You can use inclusion/exclusion:
$$\sum_{n\leq x} d(n) = \sum_{mn\le x} 1 = \sum_{m\le \sqrt{x}} \ \sum_{n\le x/m} 1 + \sum_{n\le \sqrt{x}} \ \sum_{m\le x/n} 1 - \sum_{m\le\sqrt{x}} 1 \sum_{n\le\sqrt{x}} 1.$$
Now the first two double sums are the same (with the roles of $m$ and $n$ interchanged). Hence
$$ \sum_{n\leq x} d(n) = 2 \sum_{n\le \sqrt{x}} \Big\lfloor\frac{x}{n}\Big\rfloor - \Big( \big\lfloor\sqrt{x}\big\rfloor \Big)^2$$
where $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$. Now use the fact that $\lfloor x \rfloor=x+O(1)$ to finish the proof. You will have to use the identity
$$ \sum_{n\leq x} \frac{1}{n} = \log x +\gamma + O\Big(\frac{1}{x}\Big)$$
where $\gamma$ is Euler's constant.