Let $n=25$. If I understand it correctly, each inner segment (i.e. a segment between two lattice points of unit distance which are not both on the border) has to be covered by at least one side of a unit square. Consider the $4(n-1)$ segments perpendicular to one of the sides of the big square and meeting it. A unit square can only cover two of them at a time. So at least $2(n-1)$ squares are needed to cover those. If we suppose a chessboard-like coloring with black corners, choose for that the $2(n-1)$ squares of the first & last line & column which are white. Then all the remaining inner segments make up exactly the $\frac{(n-1)^2-1}2$ white squares at the interior. So we need altogether at least $\frac{n^2-1}2=312$ squares.

spanferkel, where do you pick up the entire outer boundary?

It wasn't clear to me from the wording if the boundary is to be included. If so, obviously we need at least $4(n-1)$ unit squares to cover it, and for the remaining $(n-2)\times (n-2)$ square (the boundary of which is already covered), we can then apply my above reasoning. (Still an odd number of sides.) So altogether at least $\frac{n^2+4n+3}2=\frac{(n+1)(n+3)}2$ unit squares.

Ok, but something (arithmetic?) has gone wrong combining the two analyses: just check $n = 1$, 3, 5, for example. But once this is fixed, I don't see a proof that this is actually minimal, just a construction! Here's the proof that this construction is as good as possible: first, we certainly need the $4n - 4$ squares to cover the outer boundary. Now there are $2(n - 2)(n - 3)$ edges left to cover, including the $4(n - 3)$ that touch the inner boundary of the squares we've layed down. Each of these must be covered by a square, and each square covering these covers at most two of them and at most three total edges. Thus, after we lay $m \geq 2(n - 3)$ more squares we still have at least $2n(n - 1) - 3m$ edges left to cover. Finally, each additional square may cover at most four of these edges, so we require at least $\frac{2(n- 2)(n - 3) - 3m}{4}$ more squares. Putting this together, we have at least $(4n - 4) + m + \frac{2(n - 2)(n - 3) - 3m}{4} = \frac{2n^2 + 6n - 4 + m}{4} \geq \frac{n^2 + 4n - 5}{2}$ squares, and it's easy to see that this matches the construction.

Yes, I had an error of sign. The minimal number is indeed $\frac{(n-2)^2-1}2+4(n-1)=\frac{n-1}2(n-3+8)=\frac{n^2+4n-5}2$. Note that I had said spanferkel wrote: Then all the remaining inner segments make up exactly the $\frac{(n-1)^2-1}2$ white squares at the interior. So we need altogether at least $\frac{n^2-1}2=312$ squares. This understands, given that these inner white squares have all sides distinct, that each of them covers 4 segments, so the covering is best possible.

Yes, but it presupposes a particular arrangement for the squares you'd already added.

you are right. But I think my proof is correct if modified as follows: Use induction, starting at $n=3$. Decompose the $n\times n$ square ($n$ odd) into the central $ (n-2)\times (n-2) $ square and a frame of width $1$ around it. By induction hypothesis, we need at least $ \frac{(n-2)^{2}-1}{2} $ squares for covering the inner one (without boundary). We further need at least $2(n-1)$ squares to cover all the interior (=perpendicular) segments of the frame. And these squares are distinct, so $f(n)\ge \frac{(n-2)^{2}-1}{2} +2(n-1)= \frac{n^{2}-1}{2}$. Equivalently, decompose the $n\times n$ square into concentric frames, and look what is needed at least to cover all the interior segments of each frame, As the unit squares are distinct for each frame, we get $f(n)\ge2[(n-1)+(n-3)+...+2] = \frac{n^{2}-1}{2}$. In fact I think it is almost the same as your proof.