@AxiomaticSystem has already found the correct solution. However, as this question is tagged as logical-deduction I thought it merited a second answer which demonstrates that this solution is entirely logically deducible. Here goes!
Each step of this solution will be illustrated using the diagram provided (thick blue lines are part of the solution; thin red lines are logically impossible) and a table alongside which shows the possibilities for numbers that can be consecutive in the cycle (a cross means not possible, 'Y' means 'Yes').
First, carry out step 1 of the instructions and join all horizontal and vertical pairs of identical numbers. Furthermore, rule out all pairs of numbers which would result in more than 1 diagonal line when carrying out the OP's instruction step 2:
Note next that 1 shares a diagonal with all of 3, 4, 6, 7, 8 and 9 - if there is only one diagonal max, then 1&5 must be consecutive. Furthermore, the one diagonal MUST involve the number 1 - we can rule out all other number pairs where a diagonal does not involve 1:
Let's rule out a few more potential pairings...
- The 5-6 in the rightmost column is now isolated and cannot be connected to the boardwalk - rule out 5&6;
- Similarly, the 4-8 in the leftmost column;
- Also, we cannot have 1&3, or this would isolate the top left corner (since 1 cannot be connected to 4 or 9 as well as 3, as it is already connected to 5);
- Neither can we have 1-9 at the base of the 5th column, as to connect that to the boardwalk we would need to connect 9 both to 7 and to 5 in addition to 1 (and a number can only be connected to 2 different numbers).
Note now that if 5 connects to 0, there is no longer any way to link the left and right parts of the boardwalk. In fact, to link the two sides we will need to link 5 to 9. We now know both of 5's neighbours:
9 has one more potential connection, and its only remaining options are 2 or 7. If it is 7 then there is again no way to link the left and right parts of the boardwalk, as without being able to use the 9-2-2 connection in the top-right, the only way to link the 2-2 to the rest of the boardwalk would involve linking through 1-8 AND 1-7 - but 1 has only one potential partner left. We must therefore link 9&2 and cross out all 9-7's.
2 must now link to 8, or there is no way to link 8-8 to the boardwalk. Also, 1&7 can be ruled out, as the centre-bottom 7 requires both the 1-8 and 1-7 links, which is impossible (as previously ruled out in Step 5).
Now, only one link exists between the 3rd and 4th columns: 1-4. This must be the remaining connection with 1, and provides our one diagonal, in the top left corner.
The boardwalk is now complete, as no more connections are possible. So how to fit 3, 6, 7 and 0 into the cycle? Note that 3 cannot be adjacent to 6, or this would create a second unconnected boardwalk in the bottom left corner. This means 3 must be adjacent in the cycle to 0 and 7.
Also, 7 must connect to 6, as there are no other options available for 7. Thus we need to fit 0-3-7-6 into the cycle. Since 6 cannot be adjacent to 8 (ruled out in Step 2), it must be adjacent to 4 instead, giving us the final configuration.
Now that all is said and done (and logically deduced!) we can see the final answer is really very appropriate indeed:
The blue lines in the grid depict the Greek letter PI, the mathematical constant whose value is 3.141592... Now hold on a moment - where have we seen those digits before?! I suggest you look at the first line of numbers in the grid again...!
EDIT: In fact, as the even more astute (like @LannyStrack in comments) might notice, this puzzle actually contains the first twenty digits of pi, wrapped around the edge of the grid, starting top left! 3.1415926535897932384...