Labelling the rows 1 (top) to 7 (bottom) and the columns A (left) to F (right), our first step is that
A5 and F6 are clearly shaded, which means A4, A6, B5, F5, F7, E6 are unshaded. Then B6 is also shaded, which means C6 and B7 are unshaded. For connectivity of unshaded squares, now A7, C7, and E7 must be unshaded.
Next target to be shaded is
E5, with unshaded around it, and then D6, with unshaded around it. For connectivity, now C5 must be unshaded. Then D4 must be shaded, with unshaded around it.
Now we have a few sets of squares of which at least one must be shaded:
D1 and E1, C1 and C2 and C3, D1 and E2. That means if E1 is shaded, then D1 is not shaded and so E2 is shaded, contradiction. So D1 is shaded, which means C1, E1, and D2 are unshaded, which means exactly one of C2 and C3 is shaded.
Trying a similar deduction technique again:
At least one of C2 and E3 must be shaded. So either C2 is shaded, or C3 and E3 both are. In the latter case, we get the following which is impossible (draw a line from A1 to B6):
So in fact
C2 is shaded, which means B2 and C3 are unshaded, and B1 for connectivity. Now E2 must be shaded (I think), and then everything else in the top right corner must be unshaded.
B3 must be shaded, so B4 and A3 are unshaded, and A2 for connectivity. Then A1 must be shaded and we're done: