Bond dissociation energy order for C−C, N−N, O−O and F−F single bonds Data from Chemistry LibreTexts — Strength of Covalent Bonds, Table 8.8.1: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K:

$$
\begin{array}{cr}
\hline
\text{Bond} &amp; D/\pu{kJ mol-1} \\
\hline
\ce{C-C} &amp; 346 \\
\ce{N-N} &amp; \approx 167 \\
\ce{O-O} &amp; \approx 142 \\
\ce{F-F} &amp; 155 \\
\hline
\end{array}
$$

I couldn't understand the trend of bond dissociation energy of various single bonds. Why

$$D(\ce{C-C}) &gt; D(\ce{N-N}) &gt; D(\ce{F-F}) &gt; D(\ce{O-O})?$$

I had mugged up the trend the now, and according to me BDE should continuously increase while going from left to right in a period due to decrease in size and increase in ENC. Why this irregular trend?
In chemistry there is nothing like always. You have to see any element characteristics than decide its property. The case you are asking about:

Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., $\ce{N, O, F}$), which are unusually weak compared with single bonds between their larger congeners. It is likely that the $\ce{N–N, O–O,}$ and $\ce{F–F}$ single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms.
$\ce{N-N}gt \ce{F-F}$, in $\ce{F}$ there are greater number of lone pairs than in $\ce{N}$ so it experiences strong repulsion between its electrons that make it vulnerable to be broken.
$\ce{F-F} \gt \ce{O-O}$ it's so because $\ce{F}$ atom is very small that enables strong electrostatic attraction between nucleus and electrons
For further information on 3) you can follow this link.