Obtaining the P-value of a two-tailed t test The null hypothesis and the alternative one are stated, the null hypothesis is assumed to be true (reductio ad absurdum) and the value of significance level 0.05 is fixed. Finally, a two-tailed t test is conducted and the results are as follows:
Standard error: 46.8694
Degrees of freedom: 10
t: 0.18
Once at this point, the P-value can be obtained. The following reasoning was used to do so:
The P-value is the probability that a t having 10 degrees of freedom is more 'extreme' than -0.18. That is, less than -0.18 or greater than 0.18. 
Thus P-value: 0.86
Therefore null hypothesis is accepted.
Are this reasoning and result right? If not how could P-value be obtained?
 A: I'm assuming that you have $11$ pieces of data, since your degree of freedom is $10$. I'll be reffering to this data as 'sample', and its 'sample size' is $11$.
The sample size of n constitutes n pieces of information for estimating the mean of your sample and its variability. One degree of freedom is spent estimating the mean, and the remaining n-1 degrees of freedom estimate variability. Since we've already estimated the mean (the $t$ statistic does that in general, see below), I'll assume you have a sample with size of $11 \ (n=11)$.
The two-tailed t statistic of the $\mathbf{X} = (X_1, ...,X_{11})$ sample is:
$$t = t_{11-1} = \frac{EstimatedMean(\mathbf{X})-ActualMean(\mathbf{X})}{S_{11}^*}$$
Where $S_{11}^*$ is the corrected empiric variance:
$$S_{11}^* = \frac{1}{11-1} \sum_{i=1}^{11}\Bigr(X_i-EstimatedMean(\mathbf{X})\Bigl)$$
And obviously the Estimated Mean is just the regular average of your sample:
$$EstimatedMean(\mathbf{X}) = \frac{1}{11}\sum_{i=1}^{11} X_i$$
The t statistic calculates what it can from the sample provided, which only leaves the $ActualMean(\mathbf{X})$ needing to be estimated.
Here is the important part: We estimate the Actual Mean to be around the Estimated Mean, and we draw an interval centered around the Estimated Mean. The size of this interval depends on the significance level, commonly denoted as $\alpha$, in this case $\alpha = 0.05$. The (yet unkown) interval is commonly denoted as $(EstimatedMean(\mathbf{X}) - r_\alpha, EstimatedMean(\mathbf{X}) + r_\alpha)$. And we want our data to be in this interval with a $1-\alpha$ chance. This gives an equation for $r_\alpha$:
$$\Bbb{P}\Bigr(-\frac{r_\alpha}{S_{11}^*}\sqrt{11} \le t_{11-1} \le \frac{r_\alpha}{S_{11}^*}\sqrt{11}\ \Bigl) = 1-\alpha$$
Which can easily be transformed to:
$$t_{11-1}\Bigr(\frac{r_\alpha}{S_{11}^*}\sqrt{11}\Bigl) - t_{11-1}\Bigr(-\frac{r_\alpha}{S_{10}^*}\sqrt{11}\Bigl) = t_{11-1}\Bigr(\frac{r_\alpha}{S_{11}^*}\sqrt{11}\Bigl) - \biggl(1- t_{11-1}\Bigr(\frac{r_\alpha}{S_{11}^*}\sqrt{11}\Bigl)\biggr)= \\ = -1 + 2t_{10}\Bigl(\frac{r_\alpha}{S_{11}^*}\sqrt{11}\Bigr) = 1-\alpha \iff \\ \iff  t_{10}\Bigl(\frac{r_\alpha}{S_{11}^*}\sqrt{11}\Bigr) = 1 - \frac{\alpha}{2} \iff \\ \iff r_\alpha = \frac{t_{10}^{-1}\Bigl(\frac{\alpha}{2}\Bigr)S_{11}^*}{\sqrt{11}} = \frac{t_{10}^{-1}\Bigl(\frac{0.05}{2}\Bigr)S_{11}^*}{\sqrt{11}}$$
What is the p-value then? Simply put, the p-value$=\alpha$ substitution correctly solves the above equation for $r_\alpha$. This means, that the p-value is on the exact border on the interval, it's the $\alpha$, that's for which our null hypothesis is on the boarder of acceptance versus rejection.
Solving the equation for $\alpha$ gives you the p-value:
$$p-value = 2 - 2t_{10}\Bigl(\frac{r_\alpha}{S_{11}^*}\sqrt{11}\Bigr)$$
If the p-value is less than $\alpha$, then we reject the null hypothesis, if the p-value is greater than $\alpha$, then we accept the null hypothesis.
Additionally, the smaller the p-value is, the more we reject the null hypothesis, however, this is not true for the reverse. If the null hypothesis is right, then the distribution of the p-value is Uniform on $[0,1]$. That means that we have exactly an $\alpha$ chance to be wrong in acepting the null hypothesis.
Answering your original question: Your reasoning is not right. We accepted the null hypothesis because the p-value was greater than $\alpha$.
