Let $G$ be a connected claw-free graph of even order, and let $S$ be any set of vertices. If $S = \varnothing$, then $G-S$ has only one component, and it's even, so $o(G-S) = 0$, and Tutte's condition is satisfied for $S$.
Otherwise, to verify Tutte's condition, we will go through the vertices in $S$ one at a time, and keep track of how the number of vertices in $S$ we've visited so far (call this $k$) can compare to the number of different components of $G-S$ that any visited vertices are adjacent to (call this $h$).
Starting with an arbitrary vertex in $S$, it can be adjacent to at most two components of $G-S$, or else there will be a claw. So after looking at one vertex, we have $k=1$ and $h \le 2$.
Whenever possible, the next vertex $x \in S$ we look at should be adjacent to a vertex $y \in S$ we've already visited. In this case, $x$ cannot be adjacent to more than one component of $G-S$ that $y$ is not also adjacent to (otherwise there will be a claw with $x$, its neighbors in two such components, and $y$). So while $k$ increases by $1$, $h$ can increase by at most $1$.
Suppose that the remaining vertices in $S$ are not adjacent to any of the vertices $S$ we've already visited. In this case, pick a vertex $x \in S$ at the shortest distance from an already-visited vertex. The shortest path from $x$ to a visited vertex must go through a component of $G-S$ and then return to a vertex $y \in S$. That vertex $y$ must be already visited, or its shortest path to a visited vertex is a subset of $x$'s shortest path, and we would have picked $y$ instead of $x$. Now we observe that $x$ can be adjacent to at most two components of $G-S$, one of which $y$ was already adjacent to. So in this case, $k$ increases by $1$ and $h$ can increase by at most $1$.
We see that $h$ can only get ahead of $k$ by one, at the very first step. So at the end of the process, $k = |S|$ and $h \le |S|+1$. Moreover, $o(G-S) \le h$, because every odd component of $G-S$ has a neighbor in $S$, so it should have been counted in $h$ at some point. So $o(G-S) \le |S|+1$. Finally, we also know that $o(G-S) \equiv |S| \pmod 2$, because $G$ has even order: in particular, $o(G-S) = |S|+1$ is impossible. Therefore $o(G-S) \le |S|$, and Tutte's condition holds.
